Chapter 2 Inverse Trigonometric Functions (Additional Questions)

We are asked to find the principal value of $\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)$.

Let $y = \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)$.

By definition of inverse sine function, the range of the principal value of $\sin^{-1}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

We know that $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.

Since $\sin(-x) = -\sin(x)$, we have $\sin\left(-\frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}$.

The value $-\frac{\pi}{3}$ lies within the principal value range $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Therefore, the principal value of $\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)$ is $-\frac{\pi}{3}$.

The correct option is (B) $-\frac{\pi}{3}$.

The given function is $f(x) = \cos^{-1}(3x-1)$.

The domain of the inverse cosine function, $\cos^{-1}(u)$, is $-1 \le u \le 1$.

For the function $f(x) = \cos^{-1}(3x-1)$ to be defined, the argument $(3x-1)$ must lie within the interval $[-1, 1]$.

Therefore, we must satisfy the inequality:

To solve for $x$, we first add 1 to all parts of the inequality:

This simplifies to:

Next, we divide all parts of the inequality by 3:

This yields:

Thus, the domain of the function $f(x) = \cos^{-1}(3x-1)$ is the closed interval $[0, \frac{2}{3}]$.

Now, let's compare this result with the given options:

(A) $[0, 1]$

(B) $[-\frac{2}{3}, \frac{2}{3}]$

(C) $[-\frac{1}{3}, \frac{2}{3}]$

(D) $[-\frac{1}{3}, 0]$

The calculated domain is $[0, \frac{2}{3}]$. None of the provided options exactly match this domain. There appears to be an error in the options provided with the question.

However, if we must choose the closest option, or if there is a common type of error in question setting, we can observe that option (C) has the correct upper bound $\frac{2}{3}$, but an incorrect lower bound of $-\frac{1}{3}$. If the lower bound were 0, then (C) would be the correct answer.

Given that the question comes with multiple-choice options, and assuming there might be a typo in the options, we will state the derived correct domain and note the discrepancy.

The correct domain is $[0, \frac{2}{3}]$. None of the provided options are correct.

The principal value branch of the inverse tangent function, $\tan^{-1}x$, is the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$. This means that any angle lying strictly between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ is considered its principal value.

Let's examine each of the given options:

(A) $-\frac{3\pi}{4}$:

We know that $\frac{3\pi}{4} = 135^\circ$. So, $-\frac{3\pi}{4} = -135^\circ$. The interval $(-\frac{\pi}{2}, \frac{\pi}{2})$ corresponds to $(-90^\circ, 90^\circ)$. Since $-135^\circ$ is not between $-90^\circ$ and $90^\circ$, this angle is not in the principal value branch.

(B) $\frac{3\pi}{4}$:

This angle is $135^\circ$. Since $135^\circ$ is not between $-90^\circ$ and $90^\circ$, this angle is not in the principal value branch.

(C) $\frac{\pi}{2}$:

The principal value branch of $\tan^{-1}x$ is an open interval $(-\frac{\pi}{2}, \frac{\pi}{2})$. The value $\frac{\pi}{2}$ is not included in this interval because the tangent function is undefined at $\frac{\pi}{2}$. Therefore, this angle is not in the principal value branch.

(D) $-\frac{\pi}{6}$:

We have $-\frac{\pi}{6}$. We know that $\frac{\pi}{6} = 30^\circ$. So, $-\frac{\pi}{6} = -30^\circ$. The interval $(-\frac{\pi}{2}, \frac{\pi}{2})$ corresponds to $(-90^\circ, 90^\circ)$. Since $-30^\circ$ lies strictly between $-90^\circ$ and $90^\circ$, this angle is in the principal value branch of $\tan^{-1}x$.

Therefore, the angle that lies in the principal value branch of $\tan^{-1}x$ is $-\frac{\pi}{6}$.

The correct option is (D) $-\frac{\pi}{6}$.

The inverse cosecant function, $\text{cosec}^{-1}x$, is defined as the inverse of the cosecant function, $\text{cosec}(x)$.

To determine the principal value range of $\text{cosec}^{-1}x$, we consider the restricted domain of $\text{cosec}(x)$ such that it becomes a one-to-one function.

The cosecant function, $\text{cosec}(x) = \frac{1}{\sin(x)}$, is defined for all real numbers $x$ except where $\sin(x) = 0$. This means $x \ne n\pi$ for any integer $n$.

The range of $\text{cosec}(x)$ is $(-\infty, -1] \cup [1, \infty)$.

For the inverse function $\text{cosec}^{-1}(x)$, its domain is $(-\infty, -1] \cup [1, \infty)$.

The principal value branch for $\text{cosec}^{-1}(x)$ is typically chosen such that it covers the range of the cosecant function appropriately. The standard convention for the principal value range of $\text{cosec}^{-1}(x)$ is:

$$ \text{Range}(\text{cosec}^{-1}x) = \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\} $$

This range is chosen because:

• The interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is the principal value range for $\sin^{-1}(x)$.
• Since $\text{cosec}(x) = \frac{1}{\sin(x)}$, the values of $\text{cosec}^{-1}(x)$ correspond to the values of $\sin^{-1}(\frac{1}{x})$.
• The value $0$ is excluded from the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ because $\sin(0) = 0$, and $\text{cosec}(0)$ is undefined. If $\sin(x)$ is not $0$, then $\frac{1}{\sin(x)}$ will not be $0$.

Let's examine the given options:

(A) $[-\frac{\pi}{2}, \frac{\pi}{2}]$: This interval includes $0$, which is not allowed for the principal value of $\text{cosec}^{-1}x$.

(B) $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$: This interval correctly excludes $0$ and covers the principal values.

(C) $(0, \pi) - \{\frac{\pi}{2}\}$: This is the principal value range for $\cot^{-1}x$. Also, $\text{cosec}(\pi) = \text{undefined}$ and $\text{cosec}(0) = \text{undefined}$.

(D) $[0, \pi]$: This interval includes $0$ and $\pi$, where cosecant is undefined, and it doesn't cover the negative values for which $\text{cosec}^{-1}x$ is defined.

Therefore, the correct range of $\text{cosec}^{-1}x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$.

The correct option is (B) $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$.

We need to find the value of the expression $\tan(\tan^{-1}(10))$.

For any real number $x$, the composition of the tangent function and its inverse, $\tan(\tan^{-1}(x))$, is equal to $x$. This property holds true because the domain of $\tan^{-1}(x)$ is all real numbers, and the range of $\tan^{-1}(x)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$, which is within the domain of the tangent function.

In this case, we have $x = 10$. Since $10$ is a real number, it is in the domain of $\tan^{-1}(x)$.

Therefore, by the property of inverse trigonometric functions:

The value of $\tan(\tan^{-1}(10))$ is $10$.

The correct option is (A) $10$.

We are asked to evaluate $\sin^{-1}\left(\sin\left(\frac{5\pi}{6}\right)\right)$.

The expression is of the form $\sin^{-1}(\sin(x))$. The property $\sin^{-1}(\sin(x)) = x$ is only true if $x$ lies within the principal value range of $\sin^{-1}(x)$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

In this case, $x = \frac{5\pi}{6}$. Let's check if $\frac{5\pi}{6}$ falls within the principal value range $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

We know that $-\frac{\pi}{2} = -\frac{3\pi}{6}$ and $\frac{\pi}{2} = \frac{3\pi}{6}$.

Since $\frac{5\pi}{6} > \frac{3\pi}{6}$, the angle $\frac{5\pi}{6}$ is not in the principal value range $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

We need to find an angle $\theta$ such that $\sin(\theta) = \sin\left(\frac{5\pi}{6}\right)$ and $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

We know that $\sin(\pi - x) = \sin(x)$.

Using this identity, we can write:

Calculating the angle inside the sine function:

So, $\sin\left(\frac{5\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right)$.

Now, we can substitute this back into the original expression:

Since $\frac{\pi}{6}$ is in the principal value range $[-\frac{\pi}{2}, \frac{\pi}{2}]$, we can use the property $\sin^{-1}(\sin(x)) = x$.

Therefore,

The value of $\sin^{-1}\left(\sin\left(\frac{5\pi}{6}\right)\right)$ is $\frac{\pi}{6}$.

The correct option is (B) $\frac{\pi}{6}$.

Let's analyze the Assertion (A) and the Reason (R) separately.

Assertion (A): The domain of $\sec^{-1}(x)$ is $(-\infty, -1] \cup [1, \infty)$.

The secant function, $\sec(x)$, is defined as $\frac{1}{\cos(x)}$. For $\sec(x)$ to be defined, $\cos(x)$ must not be zero. The values of $x$ for which $\cos(x)=0$ are $x = \frac{\pi}{2} + n\pi$, where $n$ is an integer.

The range of $\cos(x)$ is $[-1, 1]$.

Therefore, the possible values of $\sec(x) = \frac{1}{\cos(x)}$ are:

• If $\cos(x) = 1$, then $\sec(x) = 1$.
• If $\cos(x) = -1$, then $\sec(x) = -1$.
• If $0 < |\cos(x)| < 1$, then $|\sec(x)| = \frac{1}{|\cos(x)|} > 1$.

So, the range of $\sec(x)$ is $(-\infty, -1] \cup [1, \infty)$.

The domain of an inverse trigonometric function is the range of the original trigonometric function. Hence, the domain of $\sec^{-1}(x)$ is indeed $(-\infty, -1] \cup [1, \infty)$.

Thus, Assertion (A) is true.

Reason (R): $\sec \theta = \frac{1}{\cos \theta}$, and $\cos \theta$ lies in $[-1, 1]$.

This statement correctly defines the secant function in terms of the cosine function and correctly states the range of the cosine function.

Thus, Reason (R) is true.

Relationship between Assertion (A) and Reason (R):

The reason explains why the domain of $\sec^{-1}(x)$ is $(-\infty, -1] \cup [1, \infty)$. Because $\sec \theta$ can only take values that are reciprocals of numbers in $[-1, 1]$ (excluding $0$), the possible values of $\sec \theta$ are $\ge 1$ or $\le -1$. Consequently, for $\sec^{-1}(x)$ to be defined, its argument $x$ must be in the set $(-\infty, -1] \cup [1, \infty)$. The reason directly leads to the assertion.

Therefore, Reason (R) is the correct explanation for Assertion (A).

The correct option is (A) Both A and R are true and R is the correct explanation of A.

The graph of $y = \cos^{-1}x$ consists of all points $(x, y)$ such that $y = \cos^{-1}x$. This means that $\cos(y) = x$, and $y$ must be in the principal value range of $\cos^{-1}x$, which is $[0, \pi]$.

We need to check each of the given points to see if they satisfy the condition $y = \cos^{-1}x$ or equivalently, $\cos(y) = x$, with $y \in [0, \pi]$.

(A) $(0, \frac{\pi}{2})$

Here, $x = 0$ and $y = \frac{\pi}{2}$. We check if $\cos(\frac{\pi}{2}) = 0$.

We know that $\cos(\frac{\pi}{2}) = 0$. Also, $\frac{\pi}{2}$ is in the range $[0, \pi]$.

So, the point $(0, \frac{\pi}{2})$ lies on the graph of $y = \cos^{-1}x$.

(B) $(\frac{1}{2}, \frac{\pi}{6})$

Here, $x = \frac{1}{2}$ and $y = \frac{\pi}{6}$. We check if $\cos(\frac{\pi}{6}) = \frac{1}{2}$.

We know that $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$, not $\frac{1}{2}$. Therefore, this point does not lie on the graph.

(C) $(1, \pi)$

Here, $x = 1$ and $y = \pi$. We check if $\cos(\pi) = 1$.

We know that $\cos(\pi) = -1$, not $1$. Therefore, this point does not lie on the graph.

(D) $(-1, -\pi)$

Here, $x = -1$ and $y = -\pi$. The principal value range for $\cos^{-1}x$ is $[0, \pi]$. Since $-\pi$ is not in this range, this point cannot lie on the graph of $y = \cos^{-1}x$, even if $\cos(-\pi) = -1$.

Based on the analysis, only the point $(0, \frac{\pi}{2})$ satisfies the conditions for being on the graph of $y = \cos^{-1}x$.

The correct option is (A) $(0, \frac{\pi}{2})$.

The inverse secant function, $\sec^{-1}x$, is the inverse of the secant function, $\sec(x)$.

The secant function is defined as $\sec(x) = \frac{1}{\cos(x)}$.

For $\sec(x)$ to be defined, the denominator, $\cos(x)$, cannot be zero. The values of $x$ for which $\cos(x) = 0$ are $x = \frac{\pi}{2} + n\pi$, where $n$ is an integer.

The range of the cosine function is $[-1, 1]$.

Therefore, the possible values of $\sec(x) = \frac{1}{\cos(x)}$ are:

• When $\cos(x) = 1$, $\sec(x) = \frac{1}{1} = 1$.
• When $\cos(x) = -1$, $\sec(x) = \frac{1}{-1} = -1$.
• When $0 < |\cos(x)| < 1$, then $|\sec(x)| = \frac{1}{|\cos(x)|} > 1$.

This implies that the range of the secant function is $(-\infty, -1] \cup [1, \infty)$.

The domain of an inverse trigonometric function is the range of the corresponding trigonometric function.

Therefore, the domain of $\sec^{-1}(x)$ is the set of all real numbers $x$ such that $x$ is in the range of $\sec(\theta)$, which is $(-\infty, -1] \cup [1, \infty)$.

Let's review the options:

(A) $(-1, 1)$: This interval is where $\sec(x)$ is not defined for the principal branch, and where $\cos(x)$ is between $-1$ and $1$ (exclusive of endpoints). This is not the domain of $\sec^{-1}(x)$.

(B) $[-1, 1]$: This is the range of $\cos(x)$, not the domain of $\sec^{-1}(x)$.

(C) $(-\infty, -1] \cup [1, \infty)$: This is precisely the set of values that $\sec(x)$ can take, and therefore the domain of $\sec^{-1}(x)$.

(D) $(-\infty, \infty)$: This is the domain of many functions, but not specifically $\sec^{-1}(x)$ because values between $-1$ and $1$ are excluded.

The domain for which $\sec^{-1}x$ is defined is $(-\infty, -1] \cup [1, \infty)$.

The correct option is (C) $(-\infty, -1] \cup [1, \infty)$.

We are given the function $f(x) = \sin^{-1}(2x-5)$. We need to determine which of the given characteristics is true for this function.

Let's analyze the domain of $f(x)$.

The domain of $\sin^{-1}(u)$ is $[-1, 1]$. Therefore, for $f(x) = \sin^{-1}(2x-5)$ to be defined, the argument $(2x-5)$ must be in the interval $[-1, 1]$.

So, we have the inequality:

Add 5 to all parts of the inequality:

This simplifies to:

Divide all parts by 2:

This gives us:

So, the domain of $f(x)$ is $[2, 3]$.

Option (A) states that its domain is $[2, 3]$, which is true.

Option (D) states that its domain is $[0, 1]$, which is false.

Let's analyze the range of $f(x)$.

The range of the inverse sine function, $\sin^{-1}(u)$, is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Since the argument $(2x-5)$ spans the interval $[-1, 1]$ as $x$ spans $[2, 3]$, the output of $\sin^{-1}(2x-5)$ will span the entire range of $\sin^{-1}(u)$.

Therefore, the range of $f(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Option (B) states that its range is $[0, \pi]$. This is the range of $\cos^{-1}(x)$, not $\sin^{-1}(x)$, so it is false.

Let's analyze if $f(x)$ is an odd function.

A function $f(x)$ is odd if $f(-x) = -f(x)$ for all $x$ in its domain. A function is even if $f(-x) = f(x)$.

For $f(x) = \sin^{-1}(2x-5)$, let's consider $f(-x)$.

$f(-x) = \sin^{-1}(2(-x)-5) = \sin^{-1}(-2x-5)$.

Now let's consider $-f(x) = -\sin^{-1}(2x-5)$.

We need to check if $\sin^{-1}(-2x-5) = -\sin^{-1}(2x-5)$.

The property $\sin^{-1}(-u) = -\sin^{-1}(u)$ holds for all $u$ in the domain $[-1, 1]$.

For $f(-x)$, the argument is $-2x-5$. For $f(x)$, the argument is $2x-5$. For the function to be odd, the domain must be symmetric about $0$. The domain of $f(x)$ is $[2, 3]$, which is not symmetric about $0$. Therefore, it cannot be an odd function.

Option (C) states that it is an odd function, which is false.

Based on our analysis:

• Option (A) is true.
• Option (B) is false.
• Option (C) is false.
• Option (D) is false.

The only correct characteristic of the function $f(x) = \sin^{-1}(2x-5)$ among the given options is its domain.

The correct option is (A) Its domain is $[2, 3]$.

The principal value branch of the inverse cosine function, $\cos^{-1}x$, is the interval $[0, \pi]$. This means that any angle $y$ for which $\cos(y) = x$ and $0 \le y \le \pi$ is considered the principal value.

We need to check which of the given angles fall within the interval $[0, \pi]$.

(A) $0$

The angle $0$ is within the interval $[0, \pi]$ since $0 \le 0 \le \pi$. Therefore, $0$ lies in the principal value branch of $\cos^{-1}x$. For example, $\cos^{-1}(1) = 0$.

(B) $\frac{\pi}{4}$

The angle $\frac{\pi}{4}$ is within the interval $[0, \pi]$ since $0 \le \frac{\pi}{4} \le \pi$. Therefore, $\frac{\pi}{4}$ lies in the principal value branch of $\cos^{-1}x$. For example, $\cos^{-1}(\frac{\sqrt{2}}{2}) = \frac{\pi}{4}$.

(C) $\frac{3\pi}{4}$

The angle $\frac{3\pi}{4}$ is within the interval $[0, \pi]$ since $0 \le \frac{3\pi}{4} \le \pi$. Therefore, $\frac{3\pi}{4}$ lies in the principal value branch of $\cos^{-1}x$. For example, $\cos^{-1}(-\frac{\sqrt{2}}{2}) = \frac{3\pi}{4}$.

(D) $-\frac{\pi}{4}$

The angle $-\frac{\pi}{4}$ is not within the interval $[0, \pi]$ since $-\frac{\pi}{4} < 0$. Therefore, $-\frac{\pi}{4}$ does not lie in the principal value branch of $\cos^{-1}x$. While there is an angle whose cosine is related to $\cos(-\frac{\pi}{4})$, the principal value itself is not $-\frac{\pi}{4}$. For example, $\cos(\frac{5\pi}{4}) = \cos(-\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2}$, and $\cos^{-1}(-\frac{\sqrt{2}}{2}) = \frac{3\pi}{4}$, not $-\frac{\pi}{4}$.

The values that lie in the principal value branch of $\cos^{-1}x$ are $0$, $\frac{\pi}{4}$, and $\frac{3\pi}{4}$.

The correct options are (A) $0$, (B) $\frac{\pi}{4}$, and (C) $\frac{3\pi}{4}$.

We need to identify which of the given statements is incorrect regarding the principal values of inverse trigonometric functions.

(A) $\sin^{-1}(-\frac{1}{2}) = -\frac{\pi}{6}$

The principal value range of $\sin^{-1}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

We know that $\sin(-\frac{\pi}{6}) = -\frac{1}{2}$.

Since $-\frac{\pi}{6}$ is in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$, this statement is correct.

(B) $\cos^{-1}(-\frac{1}{2}) = \frac{2\pi}{3}$

The principal value range of $\cos^{-1}(x)$ is $[0, \pi]$.

We know that $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$.

Since $\frac{2\pi}{3}$ is in the range $[0, \pi]$, this statement is correct.

(C) $\tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$

The principal value range of $\tan^{-1}(x)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

We know that $\tan(-\frac{\pi}{3}) = -\sqrt{3}$.

Since $-\frac{\pi}{3}$ is in the range $(-\frac{\pi}{2}, \frac{\pi}{2})$, this statement is correct.

(D) $\sec^{-1}(-2) = \frac{\pi}{3}$

The principal value range of $\sec^{-1}(x)$ is $[0, \pi] - \{\frac{\pi}{2}\}$.

We need to find the angle $y$ such that $\sec(y) = -2$ and $y \in [0, \pi] - \{\frac{\pi}{2}\}$.

If $\sec(y) = -2$, then $\cos(y) = \frac{1}{\sec(y)} = \frac{1}{-2} = -\frac{1}{2}$.

We need to find $y$ such that $\cos(y) = -\frac{1}{2}$ and $y \in [0, \pi] - \{\frac{\pi}{2}\}$.

The value of $y$ for which $\cos(y) = -\frac{1}{2}$ in the interval $[0, \pi]$ is $\frac{2\pi}{3}$.

The statement claims $\sec^{-1}(-2) = \frac{\pi}{3}$. Let's check if this is true.

If $\sec^{-1}(-2) = \frac{\pi}{3}$, then $\sec(\frac{\pi}{3}) = -2$. However, $\sec(\frac{\pi}{3}) = \frac{1}{\cos(\frac{\pi}{3})} = \frac{1}{1/2} = 2$. Since $2 \ne -2$, this statement is false.

Therefore, the statement that is NOT a principal value is (D).

The correct option is (D) $\sec^{-1}(-2) = \frac{\pi}{3}$.

We need to evaluate the expression $\tan^{-1}(1) + \cos^{-1}\left(\frac{1}{2}\right)$.

First, let's find the principal value of $\tan^{-1}(1)$.

Let $y_1 = \tan^{-1}(1)$. This means $\tan(y_1) = 1$. The principal value range for $\tan^{-1}(x)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

We know that $\tan(\frac{\pi}{4}) = 1$. Since $\frac{\pi}{4}$ is in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, the principal value of $\tan^{-1}(1)$ is $\frac{\pi}{4}$.

Next, let's find the principal value of $\cos^{-1}\left(\frac{1}{2}\right)$.

Let $y_2 = \cos^{-1}\left(\frac{1}{2}\right)$. This means $\cos(y_2) = \frac{1}{2}$. The principal value range for $\cos^{-1}(x)$ is $[0, \pi]$.

We know that $\cos(\frac{\pi}{3}) = \frac{1}{2}$. Since $\frac{\pi}{3}$ is in the interval $[0, \pi]$, the principal value of $\cos^{-1}\left(\frac{1}{2}\right)$ is $\frac{\pi}{3}$.

Now, we need to add these two values:

$\tan^{-1}(1) + \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{4} + \frac{\pi}{3}$.

To add these fractions, we find a common denominator, which is 12.

$\frac{\pi}{4} + \frac{\pi}{3} = \frac{3\pi}{12} + \frac{4\pi}{12} = \frac{3\pi + 4\pi}{12} = \frac{7\pi}{12}$.

The value of $\tan^{-1}(1) + \cos^{-1}\left(\frac{1}{2}\right)$ is $\frac{7\pi}{12}$.

The correct option is (C) $\frac{7\pi}{12}$.

We need to evaluate the expression $\cos\left(\sin^{-1}\left(\frac{4}{5}\right)\right)$.

Let $\theta = \sin^{-1}\left(\frac{4}{5}\right)$. This means that $\sin(\theta) = \frac{4}{5}$.

The principal value range of $\sin^{-1}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Since $\frac{4}{5}$ is positive, $\theta$ will be in the first quadrant, i.e., $0 < \theta < \frac{\pi}{2}$.

We need to find $\cos(\theta)$. We can use the trigonometric identity $\sin^2(\theta) + \cos^2(\theta) = 1$.

Substituting the value of $\sin(\theta)$:

This gives:

Now, solve for $\cos^2(\theta)$:

$\cos^2(\theta) = \frac{25}{25} - \frac{16}{25} = \frac{9}{25}$

Taking the square root of both sides:

Since $\theta$ is in the first quadrant ($0 < \theta < \frac{\pi}{2}$), $\cos(\theta)$ must be positive.

Therefore, $\cos(\theta) = \frac{3}{5}$.

So, $\cos\left(\sin^{-1}\left(\frac{4}{5}\right)\right) = \cos(\theta) = \frac{3}{5}$.

The correct option is (A) $\frac{3}{5}$.

We are given the equation $\sin^{-1}x = y$. We need to determine the range of $y$ for the principal value branch of the inverse sine function.

The inverse sine function, denoted as $\sin^{-1}(x)$ or $\arcsin(x)$, is defined as the inverse of the sine function restricted to a specific interval so that it becomes a one-to-one function. This restricted interval is called the principal value branch.

The sine function, $\sin(x)$, is restricted to the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ to make it one-to-one. For any value of $x$ in the domain $[-1, 1]$, the value of $\sin^{-1}(x)$ is the unique angle $y$ in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $\sin(y) = x$.

Therefore, the range of the principal value branch of $\sin^{-1}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Let's examine the given options:

(A) $(-\frac{\pi}{2}, \frac{\pi}{2})$: This is an open interval, meaning the endpoints $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ are excluded. However, the range of $\sin^{-1}(x)$ includes these endpoints (e.g., $\sin^{-1}(1) = \frac{\pi}{2}$ and $\sin^{-1}(-1) = -\frac{\pi}{2}$).

(B) $[0, \pi]$: This is the principal value range for $\cos^{-1}(x)$.

(C) $[-\frac{\pi}{2}, \frac{\pi}{2}]$: This is a closed interval that includes both $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. This is the correct principal value range for $\sin^{-1}(x)$.

(D) $(0, \pi)$: This is an open interval that excludes $0$ and $\pi$. This is not the principal value range for $\sin^{-1}(x)$.

Hence, if $\sin^{-1}x = y$, the range of $y$ for the principal value branch is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

The correct option is (C) $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

We need to determine the value of $\tan^{-1}x + \tan^{-1}y$ given the conditions $x, y \in (-1, 1)$ and $xy < 1$.

We use the trigonometric identity for the sum of two tangents:

$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

Let $A = \tan^{-1}x$ and $B = \tan^{-1}y$.

Then $\tan A = x$ and $\tan B = y$.

Substituting these into the identity:

Taking the inverse tangent of both sides:

Substituting back $A = \tan^{-1}x$ and $B = \tan^{-1}y$:

This formula is valid under certain conditions for $x$ and $y$. The principal value range of $\tan^{-1}x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$. Therefore, $-\frac{\pi}{2} < \tan^{-1}x < \frac{\pi}{2}$ and $-\frac{\pi}{2} < \tan^{-1}y < \frac{\pi}{2}$.

Adding these inequalities, we get $-\pi < \tan^{-1}x + \tan^{-1}y < \pi$.

The condition $x, y \in (-1, 1)$ ensures that $\tan^{-1}x$ and $\tan^{-1}y$ are in $(-\frac{\pi}{4}, \frac{\pi}{4})$. When we add them, the sum $\tan^{-1}x + \tan^{-1}y$ will be in $(-\frac{\pi}{2}, \frac{\pi}{2})$.

The condition $xy < 1$ ensures that $1-xy \ne 0$ and that the argument $\frac{x+y}{1-xy}$ does not lead to an angle outside the primary range $(-\frac{\pi}{2}, \frac{\pi}{2})$ for which the formula holds directly without adding or subtracting $\pi$.

Since the given conditions satisfy the requirements for this identity, the formula is directly applicable.

Therefore, for $x, y \in (-1, 1)$ such that $xy < 1$, $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$.

The correct option is (A) $\tan^{-1}\left(\frac{x+y}{1-xy}\right)$.

We are asked to find the values of $x$ for which the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ holds true.

The domain of the function $\sin^{-1}x$ is $[-1, 1]$.

The domain of the function $\cos^{-1}x$ is also $[-1, 1]$.

For the sum $\sin^{-1}x + \cos^{-1}x$ to be defined, $x$ must be in the intersection of the domains of both functions.

The intersection of $[-1, 1]$ and $[-1, 1]$ is $[-1, 1]$.

The identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ is a fundamental identity in inverse trigonometry and it holds true for all values of $x$ for which both $\sin^{-1}x$ and $\cos^{-1}x$ are defined.

Therefore, the identity holds true for all $x \in [-1, 1]$.

Let's verify this. Let $y = \sin^{-1}x$. Then $\sin(y) = x$, and $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$.

We know that $\cos(\frac{\pi}{2} - y) = \sin(y)$.

So, $\cos(\frac{\pi}{2} - y) = x$.

This implies that $\frac{\pi}{2} - y = \cos^{-1}(x)$.

Rearranging, we get $y = \frac{\pi}{2} - \cos^{-1}(x)$.

Substituting back $y = \sin^{-1}x$:

Rearranging this equation, we get:

This derivation is valid as long as $y$ is in the principal value range of $\sin^{-1}(x)$ and $\frac{\pi}{2} - y$ is in the principal value range of $\cos^{-1}(x)$.

When $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$, then $-\frac{\pi}{2} \le -y \le \frac{\pi}{2}$.

Adding $\frac{\pi}{2}$ to the inequality: $\frac{\pi}{2} - \frac{\pi}{2} \le \frac{\pi}{2} - y \le \frac{\pi}{2} + \frac{\pi}{2}$.

This gives $0 \le \frac{\pi}{2} - y \le \pi$.

This range $[0, \pi]$ is exactly the principal value range of $\cos^{-1}(x)$. Therefore, the identity holds for all $x$ in the domain of $\sin^{-1}x$ and $\cos^{-1}x$.

The values of $x$ for which the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ holds true are $x \in [-1, 1]$.

The correct option is (B) $x \in [-1, 1]$.

We need to identify which of the given expressions are equal to $2\tan^{-1}x$ under the specified conditions.

Let $\theta = \tan^{-1}x$. This implies $\tan(\theta) = x$, and the principal value range for $\theta$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

We will use the double angle formulas:

• $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta}$
• $\cos(2\theta) = \frac{1-\tan^2\theta}{1+\tan^2\theta}$
• $\tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta}$

Substituting $\tan(\theta) = x$:

• $\sin(2\theta) = \frac{2x}{1+x^2}$
• $\cos(2\theta) = \frac{1-x^2}{1+x^2}$
• $\tan(2\theta) = \frac{2x}{1-x^2}$

Now, let's analyze each option:

(A) $\sin^{-1}\left(\frac{2x}{1+x^2}\right)$, for $|x| \leq 1$

If $2\theta = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$, then $\sin(2\theta) = \frac{2x}{1+x^2}$. This identity holds true for $2\theta$. The range of $\sin^{-1}(u)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. So, we need $2\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, which means $\theta \in [-\frac{\pi}{4}, \frac{\pi}{4}]$.

Since $\tan(\theta) = x$, this means $x \in [\tan(-\frac{\pi}{4}), \tan(\frac{\pi}{4})] = [-1, 1]$.

Thus, for $|x| \le 1$, $2\tan^{-1}x = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$. This statement is correct.

(B) $\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, for $x \geq 0$

If $2\theta = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, then $\cos(2\theta) = \frac{1-x^2}{1+x^2}$. This identity holds true for $2\theta$. The range of $\cos^{-1}(u)$ is $[0, \pi]$. So, we need $2\theta \in [0, \pi]$, which means $\theta \in [0, \frac{\pi}{2}]$.

Since $\tan(\theta) = x$, and $\theta \in [0, \frac{\pi}{2})$, this means $x \in [0, \infty)$. The condition given is $x \ge 0$. For $x=0$, $2\tan^{-1}(0) = 0$ and $\cos^{-1}(\frac{1-0}{1+0}) = \cos^{-1}(1) = 0$. For $x > 0$, $\tan(\theta) > 0$, so $\theta \in (0, \frac{\pi}{2})$, which means $2\theta \in (0, \pi)$. This matches the range of $\cos^{-1}$.

Thus, for $x \ge 0$, $2\tan^{-1}x = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$. This statement is correct.

(C) $\tan^{-1}\left(\frac{2x}{1-x^2}\right)$, for $|x| < 1$

If $2\theta = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$, then $\tan(2\theta) = \frac{2x}{1-x^2}$. This identity holds true for $2\theta$. The range of $\tan^{-1}(u)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$. So, we need $2\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$, which means $\theta \in (-\frac{\pi}{4}, \frac{\pi}{4})$.

Since $\tan(\theta) = x$, this means $x \in (\tan(-\frac{\pi}{4}), \tan(\frac{\pi}{4})) = (-1, 1)$.

Thus, for $|x| < 1$, $2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$. This statement is correct.

(D) $\sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, for $|x| \leq 1$

The expression $\frac{1-x^2}{1+x^2}$ is equal to $\cos(2\theta)$, not $\sin(2\theta)$. Therefore, $\sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ is not equal to $2\tan^{-1}x$ in general.

For example, let $x=1$. Then $2\tan^{-1}(1) = 2(\frac{\pi}{4}) = \frac{\pi}{2}$.

The expression becomes $\sin^{-1}\left(\frac{1-1^2}{1+1^2}\right) = \sin^{-1}\left(\frac{0}{2}\right) = \sin^{-1}(0) = 0$.

Since $\frac{\pi}{2} \ne 0$, this statement is incorrect.

The identities that hold true are given in options (A), (B), and (C).

The correct options are (A) $\sin^{-1}\left(\frac{2x}{1+x^2}\right)$, for $|x| \leq 1$, (B) $\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, for $x \geq 0$, and (C) $\tan^{-1}\left(\frac{2x}{1-x^2}\right)$, for $|x| < 1$.

We need to express $\sin^{-1}x$ in terms of other inverse trigonometric functions for $x \in [0, 1]$.

Let $y = \sin^{-1}x$. Since $x \in [0, 1]$, the principal value of $y$ will be in the range $[0, \frac{\pi}{2}]$.

From $y = \sin^{-1}x$, we have $\sin(y) = x$. Since $y \in [0, \frac{\pi}{2}]$, we can consider a right-angled triangle where the angle is $y$.

In a right-angled triangle, if $\sin(y) = \frac{\text{opposite}}{\text{hypotenuse}}$, we can set the opposite side to $x$ and the hypotenuse to $1$.

Using the Pythagorean theorem, the adjacent side can be found: $adjacent^2 + opposite^2 = hypotenuse^2$.

$adjacent^2 + x^2 = 1^2$

$adjacent^2 = 1 - x^2$

$adjacent = \sqrt{1-x^2}$ (Since $y \in [0, \frac{\pi}{2}]$, the adjacent side is positive).

Now we can express $\tan(y)$ and $\cot(y)$ in terms of $x$:

$\tan(y) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{\sqrt{1-x^2}}$

$\cot(y) = \frac{\text{adjacent}}{\text{opposite}} = \frac{\sqrt{1-x^2}}{x}$

Taking the inverse tangent and inverse cotangent:

$y = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$

$y = \cot^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)$

Since $y = \sin^{-1}x$, we have:

$\sin^{-1}x = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$

$\sin^{-1}x = \cot^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)$

We also know that $\cot^{-1}(z) = \tan^{-1}(\frac{1}{z})$ for $z > 0$. So, $\cot^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right) = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ provided $\frac{\sqrt{1-x^2}}{x} > 0$. This is true for $x \in (0, 1)$. For $x=0$, $\sin^{-1}(0)=0$, $\tan^{-1}(0/\sqrt{1})=0$. For $x=1$, $\sin^{-1}(1)=\pi/2$, but $\tan^{-1}(1/\sqrt{0})$ is undefined. However, the expression $\frac{x}{\sqrt{1-x^2}}$ approaches infinity as $x \to 1^-$, and $\tan^{-1}$ of infinity is $\pi/2$. So the equality holds.

We need to check the given options:

(A) $\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$: This matches our derived expression.

(B) $\cot^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$: This is not directly matching, as we got $\cot^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)$.

(C) $\tan^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)$: This is equal to $\cot^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$, which is not what we derived for $\tan^{-1}$ of the adjacent/opposite ratio.

(D) $\tan^{-1}\left(\frac{1}{x}\right)$: This is $\cot^{-1}(x)$, which is not $\sin^{-1}(x)$ for the given domain.

Therefore, for $x \in [0, 1]$, $\sin^{-1}x$ can be expressed as $\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$.

The correct option is (A) $\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$.

We need to evaluate the sum $\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right)$.

We use the identity for the sum of two inverse tangents:

$\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, provided $xy < 1$.

Here, $x = \frac{1}{2}$ and $y = \frac{1}{3}$.

First, let's check the condition $xy < 1$.

$xy = \left(\frac{1}{2}\right) \times \left(\frac{1}{3}\right) = \frac{1}{6}$.

Since $\frac{1}{6} < 1$, the condition is satisfied, and we can directly apply the formula.

Substitute the values of $x$ and $y$ into the formula:

Now, we simplify the expression inside the $\tan^{-1}$ function:

Numerator: $\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$.

Denominator: $1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{5}{6}$.

So, the expression becomes:

Simplifying the fraction inside the $\tan^{-1}$:

Therefore, the sum is:

We know that $\tan^{-1}(1)$ is the angle whose tangent is 1. The principal value is $\frac{\pi}{4}$.

The value of $\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right)$ is $\tan^{-1}(1)$, which is equal to $\frac{\pi}{4}$.

Looking at the options, both (C) and (D) represent the final value.

Option (D) is $\tan^{-1}(1)$.

Option (C) is $\frac{\pi}{4}$.

Since $\tan^{-1}(1) = \frac{\pi}{4}$, both options represent the same value. However, in multiple-choice questions, if a simplified numerical value is an option, it is often preferred.

Let's re-examine the options provided.

(A) $\tan^{-1}\left(\frac{5}{6}\right)$: Incorrect.

(B) $\tan^{-1}\left(\frac{1}{6}\right)$: Incorrect.

(C) $\frac{\pi}{4}$: This is the numerical value.

(D) $\tan^{-1}(1)$: This is an expression that evaluates to $\frac{\pi}{4}$.

In many contexts, if an expression simplifies to a known angle, providing the angle is the expected answer.

The correct option is (C) $\frac{\pi}{4}$ (which is also equal to $\tan^{-1}(1)$).

We need to simplify the expression $\tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right)$ for $x \in (-\frac{\pi}{4}, \frac{\pi}{4})$.

To simplify the argument of the $\tan^{-1}$ function, we can divide the numerator and the denominator by $\cos x$ (since $x \in (-\frac{\pi}{4}, \frac{\pi}{4})$, $\cos x \ne 0$):

This simplifies to:

We know the tangent addition formula: $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.

We can rewrite $\frac{1 - \tan x}{1 + \tan x}$ by noting that $1 = \tan(\frac{\pi}{4})$:

This expression is in the form of $\tan(A - B)$, where $A = \frac{\pi}{4}$ and $B = x$.

So, $\frac{1 - \tan x}{1 + \tan x} = \tan\left(\frac{\pi}{4} - x\right)$.

Now, we can substitute this back into the original expression:

The identity $\tan^{-1}(\tan(z)) = z$ holds when $z$ is in the principal value range of $\tan^{-1}$, which is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

We are given that $x \in (-\frac{\pi}{4}, \frac{\pi}{4})$.

Let's find the range of $\frac{\pi}{4} - x$:

Since $-\frac{\pi}{4} < x < \frac{\pi}{4}$, multiplying by -1 reverses the inequalities: $\frac{\pi}{4} > -x > -\frac{\pi}{4}$, which can be written as $-\frac{\pi}{4} < -x < \frac{\pi}{4}$.

Now, add $\frac{\pi}{4}$ to all parts of the inequality:

$\frac{\pi}{4} - \frac{\pi}{4} < \frac{\pi}{4} - x < \frac{\pi}{4} + \frac{\pi}{4}$

$0 < \frac{\pi}{4} - x < \frac{\pi}{2}$

The interval $(0, \frac{\pi}{2})$ is within the principal value range $(-\frac{\pi}{2}, \frac{\pi}{2})$ of $\tan^{-1}$. Therefore, we can apply the identity $\tan^{-1}(\tan(z)) = z$.

So, $\tan^{-1}\left(\tan\left(\frac{\pi}{4} - x\right)\right) = \frac{\pi}{4} - x$.

The simplified expression is $\frac{\pi}{4} - x$.

The correct option is (A) $\frac{\pi}{4} - x$.

Let's analyze Assertion (A) and Reason (R) separately.

Assertion (A): $\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ if $x, y > 0$ and $xy > 1$.

Let $A = \tan^{-1}x$ and $B = \tan^{-1}y$. Since $x, y > 0$, $A$ and $B$ are in the interval $(0, \frac{\pi}{2})$.

The sum $A+B$ will be in the interval $(0, \pi)$.

We know that $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{x+y}{1-xy}$.

If $xy > 1$ and $x, y > 0$, then $1-xy < 0$. This means $\frac{x+y}{1-xy}$ is negative (since $x+y>0$).

If $\tan(A+B)$ is negative, and $A+B \in (0, \pi)$, then $A+B$ must be in the second quadrant, i.e., $A+B \in (\frac{\pi}{2}, \pi)$.

The principal value of $\tan^{-1}\left(\frac{x+y}{1-xy}\right)$ will lie in $(-\frac{\pi}{2}, 0)$ because the argument is negative.

Let $Z = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$. Then $\tan(Z) = \frac{x+y}{1-xy}$.

We have $\tan(A+B) = \tan(Z)$. Since $A+B \in (\frac{\pi}{2}, \pi)$ and $Z \in (-\frac{\pi}{2}, 0)$, we know that $A+B$ and $Z$ differ by a multiple of $\pi$. Specifically, $A+B = Z + \pi$.

Rearranging this, we get $A+B = \pi + Z$.

Substituting back, $\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)$.

Thus, Assertion (A) is true.

Reason (R): The formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ is only valid when $xy < 1$.

This is the principal formula for the sum of inverse tangents. This formula gives a result in the range $(-\frac{\pi}{2}, \frac{\pi}{2})$.

If $x, y > 0$ and $xy > 1$, then $1-xy < 0$, and $\frac{x+y}{1-xy} < 0$. The principal value of $\tan^{-1}\left(\frac{x+y}{1-xy}\right)$ will be in $(-\frac{\pi}{2}, 0)$. However, if $x, y > 0$, then $\tan^{-1}x \in (0, \frac{\pi}{2})$ and $\tan^{-1}y \in (0, \frac{\pi}{2})$. Their sum $\tan^{-1}x + \tan^{-1}y$ will be in $(0, \pi)$. If $xy > 1$, the sum is actually in $(\frac{\pi}{2}, \pi)$. Therefore, the direct formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ is not valid in this case because the principal value of the right side will be negative, while the left side is positive.

Thus, Reason (R) is true, as the direct application of that formula assumes the sum is within the principal range $(-\frac{\pi}{2}, \frac{\pi}{2})$, which is not the case when $xy > 1$ and $x,y>0$.

Relationship between Assertion (A) and Reason (R):

Assertion (A) provides the correct formula for the case when $x, y > 0$ and $xy > 1$. Reason (R) correctly states that the simpler formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ is valid only when $xy < 1$. The reason explains why a different formula (as stated in Assertion A) is needed when $xy > 1$. The condition $xy > 1$ necessitates an adjustment (adding $\pi$) to the principal value obtained from the basic formula. Therefore, Reason (R) helps to understand why Assertion (A) is true.

Hence, both A and R are true, and R is the correct explanation of A.

The correct option is (A) Both A and R are true and R is the correct explanation of A.

We need to check the equivalence of each given expression to $\sin^{-1}x$ under the specified conditions.

(A) $\cos^{-1}\sqrt{1-x^2}$, for $x \in [0, 1]$

Let $y = \sin^{-1}x$. Since $x \in [0, 1]$, $y \in [0, \frac{\pi}{2}]$.

Then $\sin(y) = x$. For $y \in [0, \frac{\pi}{2}]$, we have $\cos(y) = \sqrt{1-\sin^2(y)} = \sqrt{1-x^2}$.

Since $y \in [0, \frac{\pi}{2}]$, and the range of $\cos^{-1}$ is $[0, \pi]$, $y$ is in the principal value range of $\cos^{-1}$.

Therefore, $y = \cos^{-1}(\cos(y)) = \cos^{-1}(\sqrt{1-x^2})$.

So, $\sin^{-1}x = \cos^{-1}\sqrt{1-x^2}$ for $x \in [0, 1]$.

Thus, option (A) is correct.

(B) $\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$, for $x \in (-1, 1)$

Let $y = \sin^{-1}x$. If $x \in (-1, 1)$, then $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$.

We consider a right-angled triangle with angle $y$, opposite side $x$, and hypotenuse $1$. The adjacent side is $\sqrt{1-x^2}$ (which is positive for $x \in (-1, 1)$).

Then $\tan(y) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{\sqrt{1-x^2}}$.

Since $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$, $y$ is in the principal value range of $\tan^{-1}$.

Therefore, $y = \tan^{-1}(\tan(y)) = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$.

So, $\sin^{-1}x = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ for $x \in (-1, 1)$.

Thus, option (B) is correct.

(C) $\text{cosec}^{-1}\left(\frac{1}{x}\right)$, for $x \in [-1, 1], x \neq 0$

Let $y = \sin^{-1}x$. Then $\sin(y) = x$.

Consider $\text{cosec}^{-1}\left(\frac{1}{x}\right)$. Let $z = \text{cosec}^{-1}\left(\frac{1}{x}\right)$.

This means $\text{cosec}(z) = \frac{1}{x}$, or $\sin(z) = x$.

For $\text{cosec}^{-1}\left(\frac{1}{x}\right)$ to be defined, $\frac{1}{x}$ must be in the domain of $\text{cosec}^{-1}$, which is $(-\infty, -1] \cup [1, \infty)$. This implies $| \frac{1}{x} | \ge 1$, so $|x| \le 1$. Also $x \ne 0$ because $\frac{1}{x}$ would be undefined.

The principal value range of $\sin^{-1}x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

The principal value range of $\text{cosec}^{-1}u$ is $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$.

If $\sin(y) = x$, and $\sin(z) = x$, and both $y$ and $z$ are in the principal value ranges of $\sin^{-1}$ and $\text{cosec}^{-1}$ respectively, then $y=z$. Since the principal value ranges are the same (except for the exclusion of 0 in cosec$^{-1}$'s range, which is handled by $x \ne 0$), this equivalence holds.

So, $\sin^{-1}x = \text{cosec}^{-1}\left(\frac{1}{x}\right)$ for $x \in [-1, 1], x \neq 0$.

Thus, option (C) is correct.

(D) $-\sin^{-1}(-x)$, for $x \in [-1, 1]$

We know that $\sin^{-1}(-u) = -\sin^{-1}(u)$.

Let $u = -x$. Then $\sin^{-1}(-(-x)) = \sin^{-1}(x)$.

So, $\sin^{-1}(x) = \sin^{-1}(-(-x))$.

Applying the property $\sin^{-1}(-u) = -\sin^{-1}(u)$, we get:

$\sin^{-1}(x) = - \sin^{-1}(-(-x))$ if $-(-x)$ is in the domain of $\sin^{-1}$ and $-\sin^{-1}(-(-x))$ is in the range.

Let's use the property directly: $-\sin^{-1}(-x)$.

Let $y = \sin^{-1}x$. We want to see if $y = -\sin^{-1}(-x)$.

Consider $-\sin^{-1}(-x)$. Let $z = \sin^{-1}(-x)$. Then $\sin(z) = -x$. The range of $z$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

We know $\sin(-z) = -\sin(z) = -(-x) = x$.

So, $\sin(-z) = x$. If $-z$ is in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$, then $\sin^{-1}(x) = -z$.

Since $z \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, then $-z \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

So, $\sin^{-1}(x) = -z$.

Substituting $z = \sin^{-1}(-x)$, we get $\sin^{-1}(x) = -\sin^{-1}(-x)$.

This means that $\sin^{-1}(x)$ is an odd function.

Thus, option (D) is correct.

The expressions equivalent to $\sin^{-1}x$ are given in options (A), (B), (C), and (D).

The correct options are (A) $\cos^{-1}\sqrt{1-x^2}$, for $x \in [0, 1]$, (B) $\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$, for $x \in (-1, 1)$, (C) $\text{cosec}^{-1}\left(\frac{1}{x}\right)$, for $x \in [-1, 1], x \neq 0$, and (D) $-\sin^{-1}(-x)$, for $x \in [-1, 1]$.

We need to identify the incorrect property among the given options.

(A) $\sin^{-1}(-x) = -\sin^{-1}x$ for $x \in [-1, 1]$.

Let $y = \sin^{-1}x$. Then $\sin(y) = x$, and $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

Consider $-\sin^{-1}(-x)$. Let $z = \sin^{-1}(-x)$. Then $\sin(z) = -x$. Since $x \in [-1, 1]$, $-x \in [-1, 1]$. The range of $z$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

We know that $\sin(-z) = -\sin(z) = -(-x) = x$. So, $\sin(-z) = x$. Since $z \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, it follows that $-z \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

Therefore, $\sin^{-1}(x) = -z = -\sin^{-1}(-x)$.

This property is correct.

(B) $\cos^{-1}(-x) = \pi - \cos^{-1}x$ for $x \in [-1, 1]$.

Let $y = \cos^{-1}x$. Then $\cos(y) = x$, and $y \in [0, \pi]$.

Consider $\pi - \cos^{-1}x = \pi - y$. Let $\alpha = \pi - y$. Since $y \in [0, \pi]$, it follows that $-y \in [-\pi, 0]$, and $\pi - y \in [0, \pi]$. So $\alpha \in [0, \pi]$.

Now let's evaluate $\cos(\alpha) = \cos(\pi - y)$.

We know that $\cos(\pi - y) = -\cos(y)$.

Since $\cos(y) = x$, we have $\cos(\alpha) = -x$.

So, $\cos(\alpha) = -x$, and $\alpha \in [0, \pi]$. This means $\alpha = \cos^{-1}(-x)$.

Therefore, $\cos^{-1}(-x) = \pi - y = \pi - \cos^{-1}x$.

This property is correct.

(C) $\tan^{-1}(-x) = -\tan^{-1}x$ for $x \in \mathbb{R}$.

Let $y = \tan^{-1}x$. Then $\tan(y) = x$, and $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$.

Consider $-\tan^{-1}(-x)$. Let $z = \tan^{-1}(-x)$. Then $\tan(z) = -x$. Since $x \in \mathbb{R}$, $-x \in \mathbb{R}$. The range of $z$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

We know that $\tan(-z) = -\tan(z) = -(-x) = x$. So, $\tan(-z) = x$. Since $z \in (-\frac{\pi}{2}, \frac{\pi}{2})$, it follows that $-z \in (-\frac{\pi}{2}, \frac{\pi}{2})$.

Therefore, $\tan^{-1}(x) = -z = -\tan^{-1}(-x)$.

This property is correct.

(D) $\cot^{-1}(-x) = -\cot^{-1}x$ for $x \in \mathbb{R}$.

Let $y = \cot^{-1}x$. Then $\cot(y) = x$, and $y \in (0, \pi)$.

Consider $-\cot^{-1}x$. This would be $-y$. The range of $-y$ would be $(-\pi, 0)$.

Now consider $\cot^{-1}(-x)$. Let $z = \cot^{-1}(-x)$. Then $\cot(z) = -x$. The range of $z$ is $(0, \pi)$.

We know that $\cot(\pi - y) = -\cot(y)$.

Since $\cot(y) = x$, then $\cot(\pi - y) = -x$.

Since $y \in (0, \pi)$, $\pi - y \in (0, \pi)$. This interval is the principal value range of $\cot^{-1}$.

So, $\cot^{-1}(-x) = \pi - y = \pi - \cot^{-1}x$.

The property stated is $\cot^{-1}(-x) = -\cot^{-1}x$. This is incorrect. The correct property is $\cot^{-1}(-x) = \pi - \cot^{-1}x$.

The incorrect property is (D).

The correct option is (D) $\cot^{-1}(-x) = -\cot^{-1}x$ for $x \in \mathbb{R}$.

We need to simplify the function $g(x) = \tan^{-1}\left(\frac{1+x}{1-x}\right)$ for $x < 1$.

We can use the tangent addition formula: $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.

Let's try to manipulate the argument of the $\tan^{-1}$ function to match this form.

The argument is $\frac{1+x}{1-x}$.

We know that $1 = \tan(\frac{\pi}{4})$.

So, we can write the argument as $\frac{\tan(\frac{\pi}{4}) + x}{1 - \tan(\frac{\pi}{4}) x}$.

If we let $\tan B = x$, then $B = \tan^{-1}x$. If we let $\tan A = \tan(\frac{\pi}{4})$, then $A = \frac{\pi}{4}$.

So, the expression inside $\tan^{-1}$ is $\frac{\tan(\frac{\pi}{4}) + \tan B}{1 - \tan(\frac{\pi}{4}) \tan B}$, which is equal to $\tan(\frac{\pi}{4} + B)$.

Thus, $\frac{1+x}{1-x} = \tan\left(\frac{\pi}{4} + \tan^{-1}x\right)$.

Therefore, $g(x) = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + \tan^{-1}x\right)\right)$.

Now we need to consider the range. Let $\theta = \tan^{-1}x$. Since $x < 1$, $\theta \in (-\frac{\pi}{2}, \frac{\pi}{4})$.

Then $\frac{\pi}{4} + \theta \in (\frac{\pi}{4} - \frac{\pi}{2}, \frac{\pi}{4} + \frac{\pi}{4}) = (-\frac{\pi}{4}, \frac{\pi}{2})$.

The interval $(-\frac{\pi}{4}, \frac{\pi}{2})$ is within the principal value range of $\tan^{-1}$, which is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

So, $\tan^{-1}\left(\tan\left(\frac{\pi}{4} + \tan^{-1}x\right)\right) = \frac{\pi}{4} + \tan^{-1}x$.

Thus, $g(x) = \frac{\pi}{4} + \tan^{-1}x$.

Let's verify this by choosing a value for $x$. Let $x=0$. If $x<1$, this is valid.

$g(0) = \tan^{-1}\left(\frac{1+0}{1-0}\right) = \tan^{-1}(1) = \frac{\pi}{4}$.

Option (A): $\tan^{-1}(0) + \frac{\pi}{4} = 0 + \frac{\pi}{4} = \frac{\pi}{4}$. This matches.

Option (B): $\tan^{-1}(0) - \frac{\pi}{4} = 0 - \frac{\pi}{4} = -\frac{\pi}{4}$. This does not match.

Option (C): $\frac{\pi}{4} - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$. This also matches.

Option (D): $0 + \frac{\pi}{4} = \frac{\pi}{4}$. This also matches.

Let's try another value, $x = \frac{1}{\sqrt{3}}$. This is less than 1.

$g(\frac{1}{\sqrt{3}}) = \tan^{-1}\left(\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}\right) = \tan^{-1}\left(\frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}}\right) = \tan^{-1}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)$

Multiply numerator and denominator by $\sqrt{3}+1$: $\tan^{-1}\left(\frac{(\sqrt{3}+1)^2}{3-1}\right) = \tan^{-1}\left(\frac{3+1+2\sqrt{3}}{2}\right) = \tan^{-1}\left(\frac{4+2\sqrt{3}}{2}\right) = \tan^{-1}(2+\sqrt{3})$.

We know that $\tan(75^\circ) = \tan(\frac{5\pi}{12}) = 2+\sqrt{3}$. So $g(\frac{1}{\sqrt{3}}) = \frac{5\pi}{12}$.

Now let's check the options:

Option (A): $\tan^{-1}(\frac{1}{\sqrt{3}}) + \frac{\pi}{4} = \frac{\pi}{6} + \frac{\pi}{4} = \frac{2\pi+3\pi}{12} = \frac{5\pi}{12}$. This matches.

Option (C): $\frac{\pi}{4} - \tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{4} - \frac{\pi}{6} = \frac{3\pi-2\pi}{12} = \frac{\pi}{12}$. This does not match.

So, option (A) is correct.

The simplification of $g(x) = \tan^{-1}\left(\frac{1+x}{1-x}\right)$ for $x < 1$ is $\tan^{-1}x + \frac{\pi}{4}$.

The correct option is (A) $\tan^{-1}x + \frac{\pi}{4}$.

We need to evaluate $\sin\left(2\tan^{-1}\left(\frac{5}{12}\right)\right)$.

Let $\theta = \tan^{-1}\left(\frac{5}{12}\right)$. This means $\tan(\theta) = \frac{5}{12}$.

Since $\frac{5}{12} > 0$, $\theta$ is in the first quadrant, i.e., $0 < \theta < \frac{\pi}{2}$.

We need to find $\sin(2\theta)$. We can use the double angle formula for sine: $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta}$.

Substitute the value of $\tan(\theta) = \frac{5}{12}$ into the formula:

First, calculate the terms:

Numerator: $2 \times \frac{5}{12} = \frac{10}{12} = \frac{5}{6}$.

Denominator: $1 + \left(\frac{5}{12}\right)^2 = 1 + \frac{25}{144} = \frac{144}{144} + \frac{25}{144} = \frac{169}{144}$.

Now, substitute these back into the formula for $\sin(2\theta)$:

To divide by a fraction, we multiply by its reciprocal:

Simplify the expression:

$\sin(2\theta) = 5 \times \frac{144}{6 \times 169} = 5 \times \frac{24}{169} = \frac{120}{169}$.

So, $\sin\left(2\tan^{-1}\left(\frac{5}{12}\right)\right) = \frac{120}{169}$.

The correct option is (C) $\frac{120}{169}$.

We are given the condition $\sin^{-1}x + \sin^{-1}y = \frac{\pi}{2}$, and we need to find the value of $\cos^{-1}x + \cos^{-1}y$.

We know the identity relating inverse sine and inverse cosine: $\sin^{-1}z + \cos^{-1}z = \frac{\pi}{2}$ for $z \in [-1, 1]$.

From this identity, we can write $\sin^{-1}z = \frac{\pi}{2} - \cos^{-1}z$, and $\cos^{-1}z = \frac{\pi}{2} - \sin^{-1}z$.

Let's apply this to our given condition and the expression we need to evaluate.

From $\sin^{-1}x + \sin^{-1}y = \frac{\pi}{2}$, we can use the relationship $\sin^{-1}z = \frac{\pi}{2} - \cos^{-1}z$ for both $x$ and $y$.

Substitute $\sin^{-1}x = \frac{\pi}{2} - \cos^{-1}x$ and $\sin^{-1}y = \frac{\pi}{2} - \cos^{-1}y$ into the given equation:

Simplify the equation:

$\pi - (\cos^{-1}x + \cos^{-1}y) = \frac{\pi}{2}$

Now, we solve for $(\cos^{-1}x + \cos^{-1}y)$:

$\cos^{-1}x + \cos^{-1}y = \frac{\pi}{2}$

Alternatively, we can consider a specific case. If $\sin^{-1}x = \frac{\pi}{4}$ and $\sin^{-1}y = \frac{\pi}{4}$, then $x = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $y = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.

Then $\cos^{-1}x = \cos^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4}$ and $\cos^{-1}y = \cos^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4}$.

So, $\cos^{-1}x + \cos^{-1}y = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$.

The value of $\cos^{-1}x + \cos^{-1}y$ is $\frac{\pi}{2}$.

The correct option is (B) $\frac{\pi}{2}$.

We need to simplify the expression $\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) + \sin^{-1}\left(\frac{2x}{1+x^2}\right)$ for $x > 1$.

Let $x = \tan\theta$. Since $x > 1$, $\theta \in (\frac{\pi}{4}, \frac{\pi}{2})$.

Then $\frac{1-x^2}{1+x^2} = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos(2\theta)$.

And $\frac{2x}{1+x^2} = \frac{2\tan\theta}{1+\tan^2\theta} = \sin(2\theta)$.

So the expression becomes $\cos^{-1}(\cos(2\theta)) + \sin^{-1}(\sin(2\theta))$.

Now we need to determine the value of $2\theta$. Since $x > 1$, we have $\theta \in (\frac{\pi}{4}, \frac{\pi}{2})$.

Multiplying by 2, we get $2\theta \in (\frac{\pi}{2}, \pi)$.

We know that $\cos^{-1}(\cos(z)) = z$ only if $z \in [0, \pi]$.

Since $2\theta \in (\frac{\pi}{2}, \pi)$, it is in the range $[0, \pi]$. So, $\cos^{-1}(\cos(2\theta)) = 2\theta$.

We also know that $\sin^{-1}(\sin(z)) = z$ only if $z \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

However, $2\theta \in (\frac{\pi}{2}, \pi)$, which is not in the principal value range of $\sin^{-1}$.

We need to find an angle $z'$ in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $\sin(z') = \sin(2\theta)$.

We use the identity $\sin(z') = \sin(\pi - 2\theta)$.

Let's check the range of $\pi - 2\theta$. Since $2\theta \in (\frac{\pi}{2}, \pi)$, then $-2\theta \in (-\pi, -\frac{\pi}{2})$.

Adding $\pi$, we get $\pi - 2\theta \in (0, \frac{\pi}{2})$.

The interval $(0, \frac{\pi}{2})$ is within the principal value range of $\sin^{-1}$.

So, $\sin^{-1}(\sin(2\theta)) = \pi - 2\theta$.

Therefore, the expression simplifies to:

$\cos^{-1}(\cos(2\theta)) + \sin^{-1}(\sin(2\theta)) = 2\theta + (\pi - 2\theta) = \pi$.

The simplified expression is $\pi$.

The correct option is (B) $\pi$.

We need to evaluate each expression in Column I and match it with the corresponding value in Column II.

(i) $\tan^{-1}(\sqrt{3})$

Let $y = \tan^{-1}(\sqrt{3})$. This means $\tan(y) = \sqrt{3}$. The principal value range of $\tan^{-1}(x)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

We know that $\tan(\frac{\pi}{3}) = \sqrt{3}$. Since $\frac{\pi}{3}$ is in $(-\frac{\pi}{2}, \frac{\pi}{2})$, the value is $\frac{\pi}{3}$.

Matches with (b) $\frac{\pi}{3}$.

(ii) $\cos^{-1}(-\frac{1}{\sqrt{2}})$

Let $y = \cos^{-1}(-\frac{1}{\sqrt{2}})$. This means $\cos(y) = -\frac{1}{\sqrt{2}}$. The principal value range of $\cos^{-1}(x)$ is $[0, \pi]$.

We know that $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. Using the identity $\cos(\pi - \theta) = -\cos(\theta)$, we have $\cos(\pi - \frac{\pi}{4}) = -\cos(\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$.

$\pi - \frac{\pi}{4} = \frac{3\pi}{4}$. Since $\frac{3\pi}{4}$ is in $[0, \pi]$, the value is $\frac{3\pi}{4}$.

Matches with (a) $\frac{3\pi}{4}$.

(iii) $\sin^{-1}(\sin \frac{7\pi}{6})$

The principal value range of $\sin^{-1}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

The angle $\frac{7\pi}{6}$ is not in this range. We need to find an angle $\theta$ in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $\sin(\theta) = \sin(\frac{7\pi}{6})$.

We know $\sin(\frac{7\pi}{6}) = \sin(\pi + \frac{\pi}{6}) = -\sin(\frac{\pi}{6}) = -\frac{1}{2}$.

We need $\sin(\theta) = -\frac{1}{2}$ where $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

We know $\sin(-\frac{\pi}{6}) = -\sin(\frac{\pi}{6}) = -\frac{1}{2}$.

Since $-\frac{\pi}{6}$ is in $[-\frac{\pi}{2}, \frac{\pi}{2}]$, the value is $-\frac{\pi}{6}$.

Matches with (d) $-\frac{\pi}{6}$.

(iv) $\cos(\frac{\pi}{2} - \sin^{-1} \frac{1}{4})$

We use the identity $\cos(\frac{\pi}{2} - \alpha) = \sin(\alpha)$.

Here, $\alpha = \sin^{-1} \frac{1}{4}$.

So, $\cos(\frac{\pi}{2} - \sin^{-1} \frac{1}{4}) = \sin(\sin^{-1} \frac{1}{4})$.

We know that $\sin(\sin^{-1} z) = z$ for $z \in [-1, 1]$.

Since $\frac{1}{4}$ is in $[-1, 1]$, $\sin(\sin^{-1} \frac{1}{4}) = \frac{1}{4}$.

Matches with (c) $\frac{1}{4}$.

The correct matching is:

(i) - (b)

(ii) - (a)

(iii) - (d)

(iv) - (c)

Therefore, the sequence of matches is (b), (a), (d), (c).

We need to evaluate the expression $\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2)$.

First, let's find the value of $\tan^{-1}(\sqrt{3})$.

Let $y_1 = \tan^{-1}(\sqrt{3})$. This means $\tan(y_1) = \sqrt{3}$. The principal value range for $\tan^{-1}(x)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

We know that $\tan(\frac{\pi}{3}) = \sqrt{3}$. Since $\frac{\pi}{3}$ is in $(-\frac{\pi}{2}, \frac{\pi}{2})$, the principal value of $\tan^{-1}(\sqrt{3})$ is $\frac{\pi}{3}$.

Next, let's find the value of $\sec^{-1}(-2)$.

Let $y_2 = \sec^{-1}(-2)$. This means $\sec(y_2) = -2$. The principal value range for $\sec^{-1}(x)$ is $[0, \pi] - \{\frac{\pi}{2}\}$.

From $\sec(y_2) = -2$, we get $\cos(y_2) = \frac{1}{\sec(y_2)} = \frac{1}{-2} = -\frac{1}{2}$.

We need to find $y_2$ such that $\cos(y_2) = -\frac{1}{2}$ and $y_2 \in [0, \pi] - \{\frac{\pi}{2}\}$.

We know that $\cos(\frac{\pi}{3}) = \frac{1}{2}$. Using the identity $\cos(\pi - \theta) = -\cos(\theta)$, we have $\cos(\pi - \frac{\pi}{3}) = -\cos(\frac{\pi}{3}) = -\frac{1}{2}$.

$\pi - \frac{\pi}{3} = \frac{3\pi}{3} - \frac{\pi}{3} = \frac{2\pi}{3}$.

Since $\frac{2\pi}{3}$ is in the range $[0, \pi] - \{\frac{\pi}{2}\}$, the principal value of $\sec^{-1}(-2)$ is $\frac{2\pi}{3}$.

Now, we need to calculate $\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2)$:

$\frac{\pi}{3} - \frac{2\pi}{3}$

To subtract these fractions, we find a common denominator, which is 3.

$\frac{\pi}{3} - \frac{2\pi}{3} = \frac{\pi - 2\pi}{3} = \frac{-\pi}{3}$.

The result of the expression is $-\frac{\pi}{3}$.

The correct option is (C) $-\frac{\pi}{3}$.

Using the identity $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$, we have:

This simplifies to:

Equating the arguments:

Rearranging gives $x(3x^2 - 4) = 0$.

The solutions are $x=0$ and $x = \pm \frac{2}{\sqrt{3}}$.

However, these solutions are not among the options. Checking the options, $x=0$ is a valid solution.

Given the discrepancy, and that $x=0$ is the only confirmed solution from the options, there might be an error in the question or options.

If forced to choose based on the confirmed solution $x=0$, it appears in options (A), (B), and (C). Without further valid solutions from the options, a definitive choice cannot be made.

Assuming there's a typo and the question intends to test the basic algebraic manipulation and $x=0$ as a primary solution.

The confirmed solution is $x=0$.

Given the options, and the strong likelihood of a question error, no definitive answer can be provided from the given choices.

Let $\theta = \tan^{-1}x$. For $0 < x < 1$, we have $0 < \theta < \frac{\pi}{4}$, which implies $0 < 2\theta < \frac{\pi}{2}$.

The identities for $0 < 2\theta < \frac{\pi}{2}$ are:

• $\sin^{-1}\left(\frac{2x}{1+x^2}\right) = \sin^{-1}(\sin(2\theta)) = 2\theta$
• $\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \cos^{-1}(\cos(2\theta)) = 2\theta$
• $\tan^{-1}\left(\frac{2x}{1-x^2}\right) = \tan^{-1}(\tan(2\theta)) = 2\theta$

Summing these, $y = 2\theta + 2\theta + 2\theta = 6\theta = 6\tan^{-1}x$.

Since $6\tan^{-1}x$ is not an option, and assuming there might be a typo in the question or options, let's re-examine common identities.

It's possible the question intended a different combination or a typo in one of the terms.

If we consider the sum of only the first two terms: $\sin^{-1}\left(\frac{2x}{1+x^2}\right) + \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2\theta + 2\theta = 4\theta = 4\tan^{-1}x$. This matches option (D).

Given the options, and the strong possibility of a typo in the question, if the intended answer is among the options, it's likely that the sum of the first two terms was the focus, leading to $4\tan^{-1}x$.

Assuming a typo where the third term should not be present or is zero, the answer would be $4\tan^{-1}x$.

We are given the equation $\sin^{-1}x = 2\sin^{-1}a$, where $a$ is a constant.

The domain of $\sin^{-1}x$ is $[-1, 1]$, and its range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

The domain of $\sin^{-1}a$ is $[-1, 1]$, and its range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Let $y = \sin^{-1}a$. Then $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$.

The equation becomes $\sin^{-1}x = 2y$.

For a solution $x$ to exist, the value of $2y$ must be within the range of $\sin^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

So, we must have $-\frac{\pi}{2} \le 2y \le \frac{\pi}{2}$.

Dividing by 2, we get $-\frac{\pi}{4} \le y \le \frac{\pi}{4}$.

Since $y = \sin^{-1}a$, this means $-\frac{\pi}{4} \le \sin^{-1}a \le \frac{\pi}{4}$.

Taking the sine of all parts of the inequality:

$\sin(-\frac{\pi}{4}) \le \sin(\sin^{-1}a) \le \sin(\frac{\pi}{4})$

$-\frac{1}{\sqrt{2}} \le a \le \frac{1}{\sqrt{2}}$.

So, for a solution $x$ to exist, we must have $|a| \le \frac{1}{\sqrt{2}}$.

If $|a| \le \frac{1}{\sqrt{2}}$, then $2\sin^{-1}a$ will be in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$. For any value $Z$ in $[-\frac{\pi}{2}, \frac{\pi}{2}]$, the equation $\sin^{-1}x = Z$ has a unique solution $x = \sin(Z)$.

Therefore, if $|a| \le \frac{1}{\sqrt{2}}$, there is a unique solution for $x$, given by $x = \sin(2\sin^{-1}a)$.

If $|a| > \frac{1}{\sqrt{2}}$, then $2\sin^{-1}a$ will be outside the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$. For example, if $a = 1$, $\sin^{-1}(1) = \frac{\pi}{2}$, so $2\sin^{-1}(1) = \pi$. $\sin^{-1}x = \pi$ has no solution as the range of $\sin^{-1}x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

So, if $|a| > \frac{1}{\sqrt{2}}$, there is no solution for $x$. This matches option (B).

Let's check the other options.

(A) Unique solution for all $a \in [-1, 1]$. This is false because if $|a| > \frac{1}{\sqrt{2}}$, there is no solution.

(C) Two solutions if $|a| < \frac{1}{\sqrt{2}}$. For a given value $Z$ in the range of $\sin^{-1}$, $\sin^{-1}x=Z$ has a unique solution $x = \sin(Z)$. The value $2\sin^{-1}a$ is a single value. So there is a unique $x$. This option is false.

(D) Always has at least one solution. This is false because if $|a| > \frac{1}{\sqrt{2}}$, there is no solution.

The correct statement is that there is no solution if $|a| > \frac{1}{\sqrt{2}}$.

The correct option is (B) No solution if $|a| > \frac{1}{\sqrt{2}}$.

We need to find the value of $\cos\left(\tan^{-1}(-1)\right)$.

First, let's find the principal value of $\tan^{-1}(-1)$.

Let $y = \tan^{-1}(-1)$. This means $\tan(y) = -1$. The principal value range of $\tan^{-1}(x)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

We know that $\tan(\frac{\pi}{4}) = 1$. Since $\tan(-y) = -\tan(y)$, we have $\tan(-\frac{\pi}{4}) = -\tan(\frac{\pi}{4}) = -1$.

Since $-\frac{\pi}{4}$ is in the range $(-\frac{\pi}{2}, \frac{\pi}{2})$, the principal value of $\tan^{-1}(-1)$ is $-\frac{\pi}{4}$.

Now, we need to find the value of $\cos\left(-\frac{\pi}{4}\right)$.

We know that $\cos(-\theta) = \cos(\theta)$.

So, $\cos\left(-\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right)$.

The value of $\cos(\frac{\pi}{4})$ is $\frac{1}{\sqrt{2}}$.

Therefore, $\cos\left(\tan^{-1}(-1)\right) = \cos\left(-\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.

The correct option is (C) $\frac{1}{\sqrt{2}}$.

We are given the equation $\sin^{-1}(\frac{1}{5}) + \cos^{-1}x = \frac{\pi}{2}$.

We know the identity $\sin^{-1}z + \cos^{-1}z = \frac{\pi}{2}$ for $z \in [-1, 1]$.

Rearranging this identity, we get $\cos^{-1}z = \frac{\pi}{2} - \sin^{-1}z$.

We can rewrite the given equation by isolating $\cos^{-1}x$:

Comparing this with the identity $\cos^{-1}z = \frac{\pi}{2} - \sin^{-1}z$, we can see that if we let $z = \frac{1}{5}$, then the equation matches the identity.

Therefore, $x$ must be equal to $\frac{1}{5}$.

We need to check if $\frac{1}{5}$ is in the domain of $\cos^{-1}x$, which is $[-1, 1]$. Since $\frac{1}{5}$ is indeed between -1 and 1, this solution is valid.

The value of $x$ that satisfies the equation is $\frac{1}{5}$.

The correct option is (A) $\frac{1}{5}$.

A function $f(x)$ has a graph symmetric about the origin if it is an odd function, meaning $f(-x) = -f(x)$ for all $x$ in its domain.

Let's examine each function:

(A) $\cos^{-1}x$

The property is $\cos^{-1}(-x) = \pi - \cos^{-1}x$. Since this is not equal to $-\cos^{-1}x$, $\cos^{-1}x$ is not an odd function and its graph is not symmetric about the origin. It is symmetric about the y-axis shifted by $\pi/2$.

(B) $\sec^{-1}x$

The property is $\sec^{-1}(-x) = \pi - \sec^{-1}x$ (for $x \ge 1$). Since this is not equal to $-\sec^{-1}x$, $\sec^{-1}x$ is not an odd function.

(C) $\tan^{-1}x$

The property is $\tan^{-1}(-x) = -\tan^{-1}x$. This satisfies the condition for an odd function.

Therefore, $\tan^{-1}x$ is an odd function, and its graph is symmetric about the origin.

(D) $\cot^{-1}x$

The property is $\cot^{-1}(-x) = \pi - \cot^{-1}x$. Since this is not equal to $-\cot^{-1}x$, $\cot^{-1}x$ is not an odd function.

The function whose graph is symmetric about the origin is $\tan^{-1}x$.

The correct option is (C) $\tan^{-1}x$.

To find the intersection points of the graphs of $y = \sin^{-1}x$ and $y = \cos^{-1}x$, we need to find the values of $x$ for which $\sin^{-1}x = \cos^{-1}x$.

We know the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for $x \in [-1, 1]$.

Let $\sin^{-1}x = \cos^{-1}x$. Substitute this into the identity:

$2\sin^{-1}x = \frac{\pi}{2}$

$\sin^{-1}x = \frac{\pi}{4}$

Taking the sine of both sides:

$x = \frac{1}{\sqrt{2}}$

Now we find the corresponding $y$ value using either equation. Using $y = \sin^{-1}x$:

$y = \sin^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4}$.

Using $y = \cos^{-1}x$:

$y = \cos^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4}$.

So, the intersection point is $(\frac{1}{\sqrt{2}}, \frac{\pi}{4})$.

Let's check the given options:

(A) $(0, \frac{\pi}{2})$: At $x=0$, $\sin^{-1}(0)=0$ and $\cos^{-1}(0)=\frac{\pi}{2}$. These are not equal.

(B) $(\frac{1}{\sqrt{2}}, \frac{\pi}{4})$: At $x=\frac{1}{\sqrt{2}}$, $\sin^{-1}(\frac{1}{\sqrt{2}})=\frac{\pi}{4}$ and $\cos^{-1}(\frac{1}{\sqrt{2}})=\frac{\pi}{4}$. These are equal. So this is the intersection point.

(C) $(1, 0)$: At $x=1$, $\sin^{-1}(1)=\frac{\pi}{2}$ and $\cos^{-1}(1)=0$. These are not equal.

(D) $(\frac{\pi}{4}, \frac{1}{\sqrt{2}})$: This point is given with $x=\frac{\pi}{4}$ and $y=\frac{1}{\sqrt{2}}$, which is incorrect format for a point $(x, y)$. Even if it meant $x=\frac{\pi}{4}$, the values of $\sin^{-1}$ and $\cos^{-1}$ would not be $\frac{1}{\sqrt{2}}$.

The intersection point of the graphs of $y = \sin^{-1}x$ and $y = \cos^{-1}x$ is $(\frac{1}{\sqrt{2}}, \frac{\pi}{4})$.

The correct option is (B) $(\frac{1}{\sqrt{2}}, \frac{\pi}{4})$.

We are given that $\text{cosec}^{-1}x = \alpha$. We need to find the value of $\sin^{-1}\frac{1}{x}$ in terms of $\alpha$.

From $\text{cosec}^{-1}x = \alpha$, we have $\text{cosec}(\alpha) = x$. This means $\frac{1}{\sin(\alpha)} = x$. Therefore, $\sin(\alpha) = \frac{1}{x}$.

The domain of $\text{cosec}^{-1}x$ is $|x| \ge 1$, and its principal value range is $\alpha \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ excluding $0$.

From $\sin(\alpha) = \frac{1}{x}$, taking the inverse sine of both sides:

Now, we need to consider the principal value range of $\sin^{-1}$. The range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Since $\alpha$ is the principal value of $\text{cosec}^{-1}x$, we know that $\alpha \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ and $\alpha \ne 0$.

Therefore, $\alpha$ is already within the principal value range of $\sin^{-1}$.

So, $\sin^{-1}(\sin(\alpha)) = \alpha$.

This leads to $\alpha = \sin^{-1}(\frac{1}{x})$.

Thus, $\sin^{-1}\frac{1}{x}$ is equal to $\alpha$, with the conditions $|x| \ge 1$ and $\alpha \in [-\frac{\pi}{2}, \frac{\pi}{2}], \alpha \neq 0$.

The correct option is (A) $\alpha$, for $|x| \geq 1, \alpha \in [-\frac{\pi}{2}, \frac{\pi}{2}], \alpha \neq 0$.

Let $\alpha = \tan^{-1}(\frac{1}{5})$ and $\beta = \tan^{-1}(\frac{1}{8})$. We need to find $\tan(2\alpha + \beta)$.

Using $2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$, we get $2\tan^{-1}(\frac{1}{5}) = \tan^{-1}\left(\frac{2(1/5)}{1-(1/5)^2}\right) = \tan^{-1}\left(\frac{2/5}{24/25}\right) = \tan^{-1}\left(\frac{5}{12}\right)$.

Now, we have $\tan^{-1}(\frac{5}{12}) + \tan^{-1}(\frac{1}{8})$. Using $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$:

$\tan^{-1}\left(\frac{5/12 + 1/8}{1 - (5/12)(1/8)}\right) = \tan^{-1}\left(\frac{(10+3)/24}{(96-5)/96}\right) = \tan^{-1}\left(\frac{13/24}{91/96}\right) = \tan^{-1}\left(\frac{13}{24} \times \frac{96}{91}\right) = \tan^{-1}\left(\frac{4}{7}\right)$.

Finally, $\tan\left(\tan^{-1}\left(\frac{4}{7}\right)\right) = \frac{4}{7}$.

Since $4/7$ is not among the options, there is likely an error in the question or options.

The calculated value is $4/7$. As this is not an option, the question or options may be incorrect.

Let's analyze Assertion (A) and Reason (R) separately.

Assertion (A): The domain of $\tan^{-1}(\frac{1}{x})$ is $\mathbb{R} - \{0\}$.

The inverse tangent function, $\tan^{-1}(u)$, is defined for all real numbers $u$.

For the expression $\tan^{-1}(\frac{1}{x})$ to be defined, the argument $\frac{1}{x}$ must be a real number. This requires that the denominator, $x$, cannot be zero.

Thus, the domain of $\tan^{-1}(\frac{1}{x})$ is all real numbers except $0$, which is $\mathbb{R} - \{0\}$.

Thus, Assertion (A) is true.

Reason (R): The range of $\tan^{-1}y$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

This is the definition of the principal value range of the inverse tangent function. The output of $\tan^{-1}y$ is the angle whose tangent is $y$, and this angle must lie in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Thus, Reason (R) is true.

Relationship between Assertion (A) and Reason (R):

Assertion (A) describes the domain of $\tan^{-1}(\frac{1}{x})$. The reason (R) describes the range of the $\tan^{-1}$ function itself. While the range of $\tan^{-1}$ is a fundamental property, it does not directly explain why the domain of $\tan^{-1}(\frac{1}{x})$ excludes $0$. The reason why $x \ne 0$ is solely due to the argument $\frac{1}{x}$ being undefined at $x=0$. The range of the $\tan^{-1}$ function is not relevant to determining its domain or the domain of a function composed with it in this way.

Therefore, while both are true statements, the reason does not explain the assertion.

The correct option is (B) Both A and R are true but R is not the correct explanation of A.

Question 1. Find the principal value of $\sin^{-1}\left(-\frac{1}{\sqrt{2}}\right)$.

Let $\theta = \sin^{-1}\left(-\frac{1}{\sqrt{2}}\right)$.

By definition of the inverse sine function, the range of principal values is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

This means we are looking for an angle $\theta$ such that $\sin(\theta) = -\frac{1}{\sqrt{2}}$ and $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.

We know that $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.

Since the sine function is odd, $\sin(-\theta) = -\sin(\theta)$.

Therefore, $\sin\left(-\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}$.

The value $-\frac{\pi}{4}$ lies within the principal value range of $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ as $-\frac{\pi}{2} \leq -\frac{\pi}{4} \leq \frac{\pi}{2}$.

Thus, the principal value of $\sin^{-1}\left(-\frac{1}{\sqrt{2}}\right)$ is $\boxed{-\frac{\pi}{4}}$.

Question 2. Find the principal value of $\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$.

Let $\theta = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$.

By definition of the inverse cosine function, the range of principal values is $[0, \pi]$.

This means we are looking for an angle $\theta$ such that $\cos(\theta) = -\frac{\sqrt{3}}{2}$ and $0 \leq \theta \leq \pi$.

We know that $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$.

Since the cosine function is negative in the second quadrant, and the range of the principal value for $\cos^{-1}$ is $[0, \pi]$, we need to find an angle in this interval whose cosine is $-\frac{\sqrt{3}}{2}$.

We know that $\cos(\pi - x) = -\cos(x)$.

Therefore, $\cos\left(\pi - \frac{\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$.

Calculating the angle: $\pi - \frac{\pi}{6} = \frac{6\pi - \pi}{6} = \frac{5\pi}{6}$.

The value $\frac{5\pi}{6}$ lies within the principal value range of $[0, \pi]$ as $0 \leq \frac{5\pi}{6} \leq \pi$.

Thus, the principal value of $\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$ is $\boxed{\frac{5\pi}{6}}$.

Question 3. Find the principal value of $\tan^{-1}(-\sqrt{3})$.

Let $\theta = \tan^{-1}(-\sqrt{3})$.

By definition of the inverse tangent function, the range of principal values is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

This means we are looking for an angle $\theta$ such that $\tan(\theta) = -\sqrt{3}$ and $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$.

We know that $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$.

Since the tangent function is odd, $\tan(-\theta) = -\tan(\theta)$.

Therefore, $\tan\left(-\frac{\pi}{3}\right) = -\tan\left(\frac{\pi}{3}\right) = -\sqrt{3}$.

The value $-\frac{\pi}{3}$ lies within the principal value range of $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ as $-\frac{\pi}{2} < -\frac{\pi}{3} < \frac{\pi}{2}$.

Thus, the principal value of $\tan^{-1}(-\sqrt{3})$ is $\boxed{-\frac{\pi}{3}}$.

Question 4. Find the domain and range of the function $f(x) = \sin^{-1}x$.

The function $f(x) = \sin^{-1}x$ is the inverse of the sine function. To find its domain and range, we consider the properties of the sine function.

Domain of $f(x) = \sin^{-1}x$:

The domain of the inverse sine function is the set of all possible input values for $x$. For $\sin^{-1}x$ to be defined, the value of $x$ must be within the range of the sine function. The range of the sine function is $[-1, 1]$. Therefore, the domain of $\sin^{-1}x$ is $[-1, 1]$.

Domain: $[-1, 1]$

Range of $f(x) = \sin^{-1}x$:

The range of the inverse sine function is the set of all possible output values of $\sin^{-1}x$. This corresponds to the restricted domain of the sine function when finding its inverse. The principal value range for $\sin^{-1}x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Range: $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

In summary:

Domain of $\sin^{-1}x$: $[-1, 1]$

Range of $\sin^{-1}x$: $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Question 5. Evaluate $\sin^{-1}\left(\sin\frac{3\pi}{5}\right)$.

We need to evaluate $\sin^{-1}\left(\sin\frac{3\pi}{5}\right)$.

The range of the principal value for $\sin^{-1}(y)$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

We are given the angle $\frac{3\pi}{5}$. Let's check if this angle falls within the principal value range.

The range is approximately $[-1.57, 1.57]$.

The angle $\frac{3\pi}{5}$ is equal to $0.6 \times \pi$, which is approximately $0.6 \times 3.14159 = 1.884954$.

Since $\frac{3\pi}{5} > \frac{\pi}{2}$, the value $\frac{3\pi}{5}$ is not in the principal range of $\sin^{-1}(y)$.

We need to find an angle $\theta$ in the range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ such that $\sin(\theta) = \sin\left(\frac{3\pi}{5}\right)$.

We know that $\sin(\pi - x) = \sin(x)$.

Let's use this identity to find an equivalent angle within the principal range:

Let $x = \frac{3\pi}{5}$. Then, $\sin\left(\frac{3\pi}{5}\right) = \sin\left(\pi - \frac{3\pi}{5}\right)$.

Calculating $\pi - \frac{3\pi}{5}$:

$\pi - \frac{3\pi}{5} = \frac{5\pi}{5} - \frac{3\pi}{5} = \frac{2\pi}{5}$.

Now, let's check if $\frac{2\pi}{5}$ is within the principal value range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

$\frac{2\pi}{5} \approx \frac{2 \times 3.14159}{5} \approx \frac{6.28318}{5} \approx 1.256636$.

Since $-\frac{\pi}{2} \approx -1.57$ and $\frac{\pi}{2} \approx 1.57$, the angle $\frac{2\pi}{5}$ lies within the range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Therefore, $\sin^{-1}\left(\sin\frac{3\pi}{5}\right) = \sin^{-1}\left(\sin\frac{2\pi}{5}\right) = \frac{2\pi}{5}$.

The final answer is $\boxed{\frac{2\pi}{5}}$.

Question 6. Evaluate $\cos^{-1}\left(\cos\frac{7\pi}{6}\right)$.

We need to evaluate $\cos^{-1}\left(\cos\frac{7\pi}{6}\right)$.

The range of the principal value for $\cos^{-1}(y)$ is $[0, \pi]$.

We are given the angle $\frac{7\pi}{6}$. Let's check if this angle falls within the principal value range.

The range is $[0, \pi]$, which is approximately $[0, 3.14159]$.

The angle $\frac{7\pi}{6}$ is equal to $\frac{7}{6} \times \pi$, which is approximately $1.167 \times \pi$, and is greater than $\pi$. Thus, $\frac{7\pi}{6}$ is not in the principal range of $\cos^{-1}(y)$.

We need to find an angle $\theta$ in the range $[0, \pi]$ such that $\cos(\theta) = \cos\left(\frac{7\pi}{6}\right)$.

We know that the cosine function has a period of $2\pi$, so $\cos(x + 2\pi) = \cos(x)$. Also, $\cos(2\pi - x) = \cos(x)$.

The angle $\frac{7\pi}{6}$ is in the third quadrant. The cosine is negative in the third quadrant.

We can write $\frac{7\pi}{6}$ as $\pi + \frac{\pi}{6}$.

Using the identity $\cos(\pi + x) = -\cos(x)$:

$\cos\left(\frac{7\pi}{6}\right) = \cos\left(\pi + \frac{\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right)$.

We need to find an angle $\theta$ in $[0, \pi]$ such that $\cos(\theta) = -\cos\left(\frac{\pi}{6}\right)$.

Using the identity $\cos(\pi - x) = -\cos(x)$:

Let $x = \frac{\pi}{6}$. Then, $\cos\left(\pi - \frac{\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right)$.

Calculating $\pi - \frac{\pi}{6}$:

$\pi - \frac{\pi}{6} = \frac{6\pi}{6} - \frac{\pi}{6} = \frac{5\pi}{6}$.

Now, let's check if $\frac{5\pi}{6}$ is within the principal value range $[0, \pi]$.

Since $0 \leq \frac{5\pi}{6} \leq \pi$, it is within the range.

Therefore, $\cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \cos^{-1}\left(\cos\frac{5\pi}{6}\right) = \frac{5\pi}{6}$.

The final answer is $\boxed{\frac{5\pi}{6}}$.

To Prove: $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for $x \in [-1, 1]$.

Proof:

Let $\theta = \sin^{-1}x$. This means $\sin \theta = x$. The principal value of $\sin^{-1}x$ lies in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

We know the trigonometric identity: $\cos(\frac{\pi}{2} - \theta) = \sin \theta$.

Substituting $\sin \theta = x$, we get:

Now, we need to consider the range of $\frac{\pi}{2} - \theta$.

Since $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, we have:

Multiplying by -1 and reversing the inequality:

Adding $\frac{\pi}{2}$ to all parts of the inequality:

This simplifies to:

From equation (i), we have $\cos(\frac{\pi}{2} - \theta) = x$.

Taking the inverse cosine of both sides, we get:

We know that the principal value of $\cos^{-1}x$ lies in the interval $[0, \pi]$, and our expression $\frac{\pi}{2} - \theta$ also lies in this interval.

Substitute $\theta = \sin^{-1}x$ back into the equation:

Rearranging the terms to match the identity we need to prove:

This proves the identity for $x \in [-1, 1]$.

We need to evaluate the expression $\sin\left(\frac{\pi}{2} - \sin^{-1}\left(-\frac{1}{2}\right)\right)$.

First, let's find the value of $\sin^{-1}\left(-\frac{1}{2}\right)$.

Let $\theta = \sin^{-1}\left(-\frac{1}{2}\right)$. This means $\sin \theta = -\frac{1}{2}$.

The principal value of $\sin^{-1}x$ lies in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

We know that $\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$.

Since $-\frac{\pi}{6}$ is within the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$, we have:

So, $\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$.

Now substitute this value back into the original expression:

$\sin\left(\frac{\pi}{2} - \sin^{-1}\left(-\frac{1}{2}\right)\right) = \sin\left(\frac{\pi}{2} - \left(-\frac{\pi}{6}\right)\right)$

Simplify the expression inside the sine function:

$\frac{\pi}{2} - \left(-\frac{\pi}{6}\right) = \frac{\pi}{2} + \frac{\pi}{6}$

To add these fractions, find a common denominator, which is 6:

$\frac{3\pi}{6} + \frac{\pi}{6} = \frac{4\pi}{6}$

Simplify the fraction:

So, the expression becomes $\sin\left(\frac{2\pi}{3}\right)$.

We know that $\sin\left(\frac{2\pi}{3}\right) = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right)$.

The value of $\sin\left(\frac{\pi}{3}\right)$ is $\frac{\sqrt{3}}{2}$.

Alternatively, we can use the identity $\sin(\frac{\pi}{2} - \alpha) = \cos(\alpha)$.

In this case, $\alpha = \sin^{-1}\left(-\frac{1}{2}\right)$.

So, $\sin\left(\frac{\pi}{2} - \sin^{-1}\left(-\frac{1}{2}\right)\right) = \cos\left(\sin^{-1}\left(-\frac{1}{2}\right)\right)$.

Let $\beta = \sin^{-1}\left(-\frac{1}{2}\right)$. Then $\sin \beta = -\frac{1}{2}$ and $\beta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

We need to find $\cos \beta$. Using the identity $\sin^2 \beta + \cos^2 \beta = 1$:

$\left(-\frac{1}{2}\right)^2 + \cos^2 \beta = 1$

$\frac{1}{4} + \cos^2 \beta = 1$

$\cos^2 \beta = 1 - \frac{1}{4} = \frac{3}{4}$

$\cos \beta = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}$

Since $\beta = \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$, and $-\frac{\pi}{6}$ lies in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$, the cosine of this angle is positive.

Therefore, $\cos \beta = \frac{\sqrt{3}}{2}$.

Final Answer: The value of the expression is $\frac{\sqrt{3}}{2}$.

We need to simplify the expression $\tan^{-1}\left(\frac{\sin x}{1+\cos x}\right)$ for $x \in (-\pi, \pi)$.

We will use the following trigonometric identities:

Substitute these identities into the expression inside the $\tan^{-1}$ function:

$\frac{\sin x}{1+\cos x} = \frac{2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}{2 \cos^2 \left(\frac{x}{2}\right)}$

Now, cancel out common terms:

$\frac{\sin x}{1+\cos x} = \frac{\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)}$

This simplifies to $\tan \left(\frac{x}{2}\right)$.

So, the expression becomes $\tan^{-1}\left(\tan \left(\frac{x}{2}\right)\right)$.

We know that $\tan^{-1}(\tan \theta) = \theta$ if $\theta$ is within the principal range of $\tan^{-1}$, which is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

We are given that $x \in (-\pi, \pi)$.

Dividing the interval by 2, we get $\frac{x}{2} \in (-\frac{\pi}{2}, \frac{\pi}{2})$.

Since $\frac{x}{2}$ is within the principal range of the tangent inverse function, we can directly apply the identity:

Final Answer: The simplified expression is $\frac{x}{2}$.

The principal value branch of an inverse trigonometric function is the specific range of values for which the inverse function is defined. This ensures that the function is one-to-one.

Principal Value Branch of $\text{cosec}^{-1}x$:

The cosecant function, $\text{cosec } x$, is not one-to-one over its entire domain. To define its inverse, we restrict the domain of $\text{cosec } x$. The standard principal value branch for $\text{cosec}^{-1}x$ is defined as follows:

The domain of $\text{cosec}^{-1}x$ is $(-\infty, -1] \cup [1, \infty)$.

The range (principal value branch) of $\text{cosec}^{-1}x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \setminus \{0\}$.

This means that for any $x$ in $(-\infty, -1] \cup [1, \infty)$, the value of $\text{cosec}^{-1}x$ will be an angle $\theta$ such that $\text{cosec } \theta = x$ and $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ but $\theta \neq 0$.

Principal Value Branch of $\sec^{-1}x$:

Similarly, the secant function, $\sec x$, is not one-to-one over its entire domain. The standard principal value branch for $\sec^{-1}x$ is defined as follows:

The domain of $\sec^{-1}x$ is $(-\infty, -1] \cup [1, \infty)$.

The range (principal value branch) of $\sec^{-1}x$ is $[0, \pi] \setminus \{\frac{\pi}{2}\}$.

This means that for any $x$ in $(-\infty, -1] \cup [1, \infty)$, the value of $\sec^{-1}x$ will be an angle $\phi$ such that $\sec \phi = x$ and $\phi \in [0, \pi]$ but $\phi \neq \frac{\pi}{2}$.

We need to evaluate the expression $\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)$.

We will evaluate each term separately:

1. Evaluate $\tan^{-1}(1)$

Let $\alpha = \tan^{-1}(1)$. This means $\tan \alpha = 1$. The principal value branch for $\tan^{-1}x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

We know that $\tan\left(\frac{\pi}{4}\right) = 1$. Since $\frac{\pi}{4}$ is in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$,

2. Evaluate $\cos^{-1}\left(-\frac{1}{2}\right)$

Let $\beta = \cos^{-1}\left(-\frac{1}{2}\right)$. This means $\cos \beta = -\frac{1}{2}$. The principal value branch for $\cos^{-1}x$ is $[0, \pi]$.

We know that $\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$. Since $\frac{2\pi}{3}$ is in the interval $[0, \pi]$,

3. Evaluate $\sin^{-1}\left(-\frac{1}{2}\right)$

Let $\gamma = \sin^{-1}\left(-\frac{1}{2}\right)$. This means $\sin \gamma = -\frac{1}{2}$. The principal value branch for $\sin^{-1}x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

We know that $\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$. Since $-\frac{\pi}{6}$ is in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$,

Now, add the values of the three terms:

$\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right) = \frac{\pi}{4} + \frac{2\pi}{3} + \left(-\frac{\pi}{6}\right)$

To add these fractions, find a common denominator, which is 12:

$= \frac{3\pi}{12} + \frac{8\pi}{12} - \frac{2\pi}{12}$

$= \frac{3\pi + 8\pi - 2\pi}{12}$

$= \frac{9\pi}{12}$

Simplify the fraction:

Final Answer: The value of the expression is $\frac{3\pi}{4}$.

We need to simplify the expression $\tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right)$ for $-\frac{\pi}{4} < x < \frac{3\pi}{4}$.

To simplify the expression inside the $\tan^{-1}$ function, we can divide both the numerator and the denominator by $\cos x$ (since $\cos x \neq 0$ in the given interval, except possibly at $x = \frac{\pi}{2}$ which is covered by the condition).

Assuming $\cos x \neq 0$:

$\frac{\cos x - \sin x}{\cos x + \sin x} = \frac{\frac{\cos x}{\cos x} - \frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x} + \frac{\sin x}{\cos x}}$

$= \frac{1 - \tan x}{1 + \tan x}$

We know the tangent addition formula: $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.

We can rewrite $\frac{1 - \tan x}{1 + \tan x}$ in the form of this formula by noting that $1 = \tan\left(\frac{\pi}{4}\right)$ and $\tan x = \tan x \cdot 1 = \tan x \tan\left(\frac{\pi}{4}\right)$ (this is not strictly needed but helps visualize).

So, $\frac{1 - \tan x}{1 + \tan x} = \frac{\tan\left(\frac{\pi}{4}\right) - \tan x}{1 + \tan\left(\frac{\pi}{4}\right) \tan x}$.

This is the expansion of $\tan\left(\frac{\pi}{4} - x\right)$.

Therefore, the expression becomes:

$\tan^{-1}\left(\tan\left(\frac{\pi}{4} - x\right)\right)$

We know that $\tan^{-1}(\tan \theta) = \theta$ if $\theta$ is in the principal range of $\tan^{-1}$, which is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

We are given that $-\frac{\pi}{4} < x < \frac{3\pi}{4}$.

Let's find the range of $\frac{\pi}{4} - x$:

Multiply the inequality by -1 and reverse the signs:

Add $\frac{\pi}{4}$ to all parts of the inequality:

$\frac{\pi}{4} - \frac{3\pi}{4} < \frac{\pi}{4} - x < \frac{\pi}{4} + \frac{\pi}{4}$

$-\frac{2\pi}{4} < \frac{\pi}{4} - x < \frac{2\pi}{4}$

$-\frac{\pi}{2} < \frac{\pi}{4} - x < \frac{\pi}{2}$

Since the argument $\left(\frac{\pi}{4} - x\right)$ lies within the principal range $(-\frac{\pi}{2}, \frac{\pi}{2})$, we can directly apply the identity $\tan^{-1}(\tan \theta) = \theta$.

Final Answer: The simplified expression is $\frac{\pi}{4} - x$.

We need to evaluate the expression $\tan^{-1}( \sqrt{3} ) - \sec^{-1}(-2)$.

We will evaluate each term separately:

1. Evaluate $\tan^{-1}( \sqrt{3} )$

Let $\alpha = \tan^{-1}( \sqrt{3} )$. This means $\tan \alpha = \sqrt{3}$. The principal value branch for $\tan^{-1}x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

We know that $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$. Since $\frac{\pi}{3}$ is in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$,

2. Evaluate $\sec^{-1}(-2)$

Let $\beta = \sec^{-1}(-2)$. This means $\sec \beta = -2$. The principal value branch for $\sec^{-1}x$ is $[0, \pi] \setminus \{\frac{\pi}{2}\}$.

We know that $\sec \beta = \frac{1}{\cos \beta}$. So, $\frac{1}{\cos \beta} = -2$, which means $\cos \beta = -\frac{1}{2}$.

We know that $\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$. Since $\frac{2\pi}{3}$ is in the interval $[0, \pi]$ and $\frac{2\pi}{3} \neq \frac{\pi}{2}$,

Now, substitute these values back into the original expression:

$\tan^{-1}( \sqrt{3} ) - \sec^{-1}(-2) = \frac{\pi}{3} - \frac{2\pi}{3}$

Subtract the fractions:

$= \frac{\pi - 2\pi}{3}$

$= \frac{-\pi}{3}$

Final Answer: The value of the expression is $-\frac{\pi}{3}$.

To Prove: $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ for $x > 0, y > 0, xy < 1$.

Proof:

Let $\tan^{-1}x = \alpha$ and $\tan^{-1}y = \beta$.

From these definitions, we have:

Given that $x > 0$ and $y > 0$, the principal values for $\alpha$ and $\beta$ will be in the first quadrant.

Since $x > 0$, $\alpha \in (0, \frac{\pi}{2})$.

Since $y > 0$, $\beta \in (0, \frac{\pi}{2})$.

Therefore, the sum $\alpha + \beta$ will be in the interval $(0, \pi)$.

Now, consider the tangent addition formula:

Substitute $\tan \alpha = x$ and $\tan \beta = y$ into the formula:

Now, we need to consider the condition $xy < 1$.

Since $x > 0$ and $y > 0$, the condition $xy < 1$ implies that $1 - xy > 0$.

Also, since $\tan \alpha = x > 0$ and $\tan \beta = y > 0$, and $\alpha, \beta \in (0, \frac{\pi}{2})$, it means that $\alpha$ and $\beta$ are acute angles.

If $xy < 1$, then $\frac{x+y}{1-xy} > 0$. This means that $\alpha + \beta$ must be in the first quadrant, i.e., $0 < \alpha + \beta < \frac{\pi}{2}$.

Taking the inverse tangent of both sides of $\tan(\alpha + \beta) = \frac{x + y}{1 - xy}$:

$\alpha + \beta = \tan^{-1}\left(\frac{x + y}{1 - xy}\right)$

Substitute back $\alpha = \tan^{-1}x$ and $\beta = \tan^{-1}y$:

This proves the identity under the given conditions $x > 0, y > 0, xy < 1$.

We need to evaluate $\sin\left(\tan^{-1}\left(\frac{3}{4}\right)\right)$.

Let $\theta = \tan^{-1}\left(\frac{3}{4}\right)$. This means $\tan \theta = \frac{3}{4}$.

The principal value branch for $\tan^{-1}x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$. Since $\frac{3}{4}$ is positive, $\theta$ lies in the first quadrant, i.e., $\theta \in (0, \frac{\pi}{2})$.

We can construct a right-angled triangle where the angle is $\theta$. Since $\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4}$, we can consider the opposite side to be 3 units and the adjacent side to be 4 units.

Using the Pythagorean theorem, the hypotenuse ($h$) can be calculated:

$h^2 = (\text{opposite})^2 + (\text{adjacent})^2$

$h^2 = 3^2 + 4^2$

$h^2 = 9 + 16$

$h^2 = 25$

$h = \sqrt{25} = 5$

Now we need to find $\sin \theta$. In a right-angled triangle, $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$.

Therefore, $\sin \theta = \frac{3}{5}$.

Since $\theta = \tan^{-1}\left(\frac{3}{4}\right)$, we have:

Final Answer: The value of the expression is $\frac{3}{5}$.

We need to find the principal value of $\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)$.

Let $\theta = \sec^{-1}\left(\frac{2}{\sqrt{3}}\right)$.

By the definition of the inverse secant function, this means $\sec \theta = \frac{2}{\sqrt{3}}$.

The principal value branch for $\sec^{-1}x$ is $[0, \pi] \setminus \{\frac{\pi}{2}\}$.

We know that $\sec \theta = \frac{1}{\cos \theta}$. Therefore, $\frac{1}{\cos \theta} = \frac{2}{\sqrt{3}}$.

This implies $\cos \theta = \frac{\sqrt{3}}{2}$.

We need to find the angle $\theta$ in the interval $[0, \pi] \setminus \{\frac{\pi}{2}\}$ such that $\cos \theta = \frac{\sqrt{3}}{2}$.

We know that $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$.

Since $\frac{\pi}{6}$ lies in the interval $[0, \pi]$ and $\frac{\pi}{6} \neq \frac{\pi}{2}$, the principal value of $\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)$ is $\frac{\pi}{6}$.

Final Answer: The principal value is $\frac{\pi}{6}$.

To Prove: $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$ for all $x \in \mathbb{R}$.

Proof:

Let $\tan^{-1}x = \theta$.

The principal value branch for $\tan^{-1}x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$. So, $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$.

From this, we have $\tan \theta = x$.

We know the trigonometric identity $\cot\left(\frac{\pi}{2} - \theta\right) = \tan \theta$.

Substituting $\tan \theta = x$, we get:

Now, we need to consider the range of $\frac{\pi}{2} - \theta$.

Since $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$:

Multiply by -1 and reverse the inequalities:

Rearranging:

Add $\frac{\pi}{2}$ to all parts of the inequality:

$\frac{\pi}{2} - \frac{\pi}{2} < \frac{\pi}{2} - \theta < \frac{\pi}{2} + \frac{\pi}{2}$

$0 < \frac{\pi}{2} - \theta < \pi$

The principal value branch for $\cot^{-1}x$ is $(0, \pi)$.

From equation (i), $\cot\left(\frac{\pi}{2} - \theta\right) = x$. Since $\frac{\pi}{2} - \theta$ lies in the interval $(0, \pi)$, we can apply the inverse cotangent function.

$\frac{\pi}{2} - \theta = \cot^{-1}x$

Substitute back $\theta = \tan^{-1}x$:

Rearranging the terms:

This holds for all $x \in \mathbb{R}$.

We need to evaluate $\cos(\sec^{-1}x + \text{cosec}^{-1}x)$ for $|x| \geq 1$.

We know the identity $\sec^{-1}x + \text{cosec}^{-1}x = \frac{\pi}{2}$ for $|x| \geq 1$.

Let's prove this identity first.

Let $\sec^{-1}x = \alpha$. Then $\sec \alpha = x$, where $\alpha \in [0, \pi] \setminus \{\frac{\pi}{2}\}$.

This means $\cos \alpha = \frac{1}{x}$.

We also know that $\sin\left(\frac{\pi}{2} - \alpha\right) = \cos \alpha$.

So, $\sin\left(\frac{\pi}{2} - \alpha\right) = \frac{1}{x}$.

Since $\alpha \in [0, \pi] \setminus \{\frac{\pi}{2}\}$, then $\frac{\pi}{2} - \alpha \in [-\frac{\pi}{2}, \frac{\pi}{2}] \setminus \{0\}$.

Taking the inverse sine of both sides:

$\frac{\pi}{2} - \alpha = \sin^{-1}\left(\frac{1}{x}\right)$

Substitute $\alpha = \sec^{-1}x$:

$\frac{\pi}{2} - \sec^{-1}x = \sin^{-1}\left(\frac{1}{x}\right)$

We also know that $\text{cosec}^{-1}x = \sin^{-1}\left(\frac{1}{x}\right)$ for $|x| \geq 1$.

Therefore, $\frac{\pi}{2} - \sec^{-1}x = \text{cosec}^{-1}x$.

Rearranging this, we get $\sec^{-1}x + \text{cosec}^{-1}x = \frac{\pi}{2}$.

Now, substitute this identity into the expression we need to evaluate:

$\cos(\sec^{-1}x + \text{cosec}^{-1}x) = \cos\left(\frac{\pi}{2}\right)$

We know that $\cos\left(\frac{\pi}{2}\right) = 0$.

Final Answer: The value of the expression is 0.

We need to simplify $\tan^{-1}\left(\frac{a \cos x - b \sin x}{b \cos x + a \sin x}\right)$, given that $\frac{a}{b} \tan x > -1$.

To simplify the expression inside the $\tan^{-1}$ function, we can divide both the numerator and the denominator by $b \cos x$ (assuming $b \cos x \neq 0$).

$\frac{a \cos x - b \sin x}{b \cos x + a \sin x} = \frac{\frac{a \cos x}{b \cos x} - \frac{b \sin x}{b \cos x}}{\frac{b \cos x}{b \cos x} + \frac{a \sin x}{b \cos x}}$

$= \frac{\frac{a}{b} - \tan x}{1 + \frac{a}{b} \tan x}$

This expression is in the form of the tangent subtraction formula: $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.

We can write $\frac{a}{b}$ as $\tan \alpha$, where $\alpha = \tan^{-1}\left(\frac{a}{b}\right)$.

So, the expression becomes:

$\frac{\tan \alpha - \tan x}{1 + \tan \alpha \tan x}$

This is equal to $\tan(\alpha - x)$.

Substituting back $\alpha = \tan^{-1}\left(\frac{a}{b}\right)$, the expression inside $\tan^{-1}$ is $\tan\left(\tan^{-1}\left(\frac{a}{b}\right) - x\right)$.

So, we have $\tan^{-1}\left(\tan\left(\tan^{-1}\left(\frac{a}{b}\right) - x\right)\right)$.

We know that $\tan^{-1}(\tan \theta) = \theta$ if $\theta$ is in the principal range $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Let $\theta = \tan^{-1}\left(\frac{a}{b}\right) - x$. We need to ensure that $\theta$ falls within $(-\frac{\pi}{2}, \frac{\pi}{2})$.

The condition given is $\frac{a}{b} \tan x > -1$. This can be written as $\tan \alpha \tan x > -1$.

Consider the condition for $\tan(\alpha - x)$ to be positive, which is when $\alpha - x$ is in $(0, \frac{\pi}{2})$. This happens when $\tan \alpha > \tan x$ and $1 + \tan \alpha \tan x > 0$. Alternatively, if $\tan \alpha < \tan x$, then $\alpha - x$ would be negative, and if $1+\tan\alpha \tan x$ is positive, then $\tan(\alpha - x)$ will be negative, and $\tan^{-1}(\tan(\alpha - x)) = \alpha - x$.

The given condition $\frac{a}{b} \tan x > -1$ ensures that $1 + \frac{a}{b} \tan x \neq 0$ and it plays a role in determining the quadrant of $\tan^{-1}\left(\frac{a}{b}\right) - x$.

Assuming that $\tan^{-1}\left(\frac{a}{b}\right) - x$ falls within the principal range of $\tan^{-1}$, we have:

Final Answer: The simplified expression is $\tan^{-1}\left(\frac{a}{b}\right) - x$.

We need to find the value of $\sin^{-1}\left(\frac{1}{2}\right) + \cos^{-1}\left(\frac{1}{2}\right)$.

We can evaluate each term separately:

1. Evaluate $\sin^{-1}\left(\frac{1}{2}\right)$

Let $\alpha = \sin^{-1}\left(\frac{1}{2}\right)$. This means $\sin \alpha = \frac{1}{2}$. The principal value branch for $\sin^{-1}x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

We know that $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$. Since $\frac{\pi}{6}$ is in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$,

2. Evaluate $\cos^{-1}\left(\frac{1}{2}\right)$

Let $\beta = \cos^{-1}\left(\frac{1}{2}\right)$. This means $\cos \beta = \frac{1}{2}$. The principal value branch for $\cos^{-1}x$ is $[0, \pi]$.

We know that $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$. Since $\frac{\pi}{3}$ is in the interval $[0, \pi]$,

Now, add the values:

$\sin^{-1}\left(\frac{1}{2}\right) + \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} + \frac{\pi}{3}$

To add these fractions, find a common denominator, which is 6:

$= \frac{\pi}{6} + \frac{2\pi}{6}$

$= \frac{\pi + 2\pi}{6}$

$= \frac{3\pi}{6}$

Simplify the fraction:

Alternatively, we can use the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for $x \in [-1, 1]$.

In this case, $x = \frac{1}{2}$, which is in the domain $[-1, 1]$.

Therefore, $\sin^{-1}\left(\frac{1}{2}\right) + \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{2}$.

Final Answer: The value of the expression is $\frac{\pi}{2}$.

To Prove: $\tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$ for $xy > -1$.

Proof:

Let $\tan^{-1}x = \alpha$ and $\tan^{-1}y = \beta$.

From these definitions, we have:

The principal value branches for $\tan^{-1}x$ and $\tan^{-1}y$ are $(-\frac{\pi}{2}, \frac{\pi}{2})$. So, $-\frac{\pi}{2} < \alpha < \frac{\pi}{2}$ and $-\frac{\pi}{2} < \beta < \frac{\pi}{2}$.

This implies that $-\pi < \alpha - \beta < \pi$.

Now, consider the tangent subtraction formula:

Substitute $\tan \alpha = x$ and $\tan \beta = y$ into the formula:

We are given that $xy > -1$, which ensures that $1 + xy \neq 0$.

Taking the inverse tangent of both sides:

$\alpha - \beta = \tan^{-1}\left(\frac{x - y}{1 + xy}\right)$

Substitute back $\alpha = \tan^{-1}x$ and $\beta = \tan^{-1}y$:

This formula is generally true, but the condition $xy > -1$ is important for the principal value of $\tan^{-1}x - \tan^{-1}y$ to be equal to $\tan^{-1}\left(\frac{x-y}{1+xy}\right)$. If $x$ and $y$ are such that $\alpha - \beta$ falls outside the principal range $(-\frac{\pi}{2}, \frac{\pi}{2})$, we might need to add or subtract $\pi$. However, for the stated condition $xy > -1$, this equality generally holds for the principal values.

We need to find the principal value of $\text{cosec}^{-1}(- \sqrt{2})$.

Let $\theta = \text{cosec}^{-1}(- \sqrt{2})$.

By the definition of the inverse cosecant function, this means $\text{cosec } \theta = -\sqrt{2}$.

The principal value branch for $\text{cosec}^{-1}x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}] \setminus \{0\}$.

We know that $\text{cosec } \theta = \frac{1}{\sin \theta}$. Therefore, $\frac{1}{\sin \theta} = -\sqrt{2}$.

This implies $\sin \theta = -\frac{1}{\sqrt{2}}$.

We need to find the angle $\theta$ in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}] \setminus \{0\}$ such that $\sin \theta = -\frac{1}{\sqrt{2}}$.

We know that $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$. Since the sine function is odd, $\sin(-\phi) = -\sin(\phi)$.

Therefore, $\sin\left(-\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}$.

Since $-\frac{\pi}{4}$ lies in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}] \setminus \{0\}$, the principal value of $\text{cosec}^{-1}(- \sqrt{2})$ is $-\frac{\pi}{4}$.

Final Answer: The principal value is $-\frac{\pi}{4}$.

Let $\theta = \tan^{-1}x$.

Then, $x = \tan \theta$.

We know that $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta}$.

Taking the inverse sine of both sides:

Substituting $\theta = \tan^{-1}x$ back:

We also know that $\cos(2\theta) = \frac{1-\tan^2\theta}{1+\tan^2\theta}$.

Taking the inverse cosine of both sides:

Substituting $\theta = \tan^{-1}x$ back:

We also know that $\tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta}$.

Taking the inverse tangent of both sides:

Substituting $\theta = \tan^{-1}x$ back:

Conditions under which these identities hold:

The principal value of $\tan^{-1}x$ lies in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$. Therefore, $-\pi < 2\theta < \pi$.

For identity (a): $2\tan^{-1}x = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$

This identity holds when $-\frac{\pi}{2} \le 2\theta \le \frac{\pi}{2}$, which means $-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}$.

If $x > 1$, then $\frac{\pi}{4} < \theta < \frac{\pi}{2}$, so $\frac{\pi}{2} < 2\theta < \pi$. In this case, $\sin(2\theta) = \frac{2x}{1+x^2}$, so $2\theta = \pi - \sin^{-1}\left(\frac{2x}{1+x^2}\right)$, which means $2\tan^{-1}x = \pi - \sin^{-1}\left(\frac{2x}{1+x^2}\right)$.

If $x < -1$, then $-\frac{\pi}{2} < \theta < -\frac{\pi}{4}$, so $-\pi < 2\theta < -\frac{\pi}{2}$. In this case, $\sin(2\theta) = \frac{2x}{1+x^2}$, so $2\theta = -\pi - \sin^{-1}\left(\frac{2x}{1+x^2}\right)$, which means $2\tan^{-1}x = -\pi - \sin^{-1}\left(\frac{2x}{1+x^2}\right)$.

Condition for (a): $-1 \le x \le 1$.

For identity (b): $2\tan^{-1}x = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

This identity holds when $0 \le 2\theta \le \pi$, which means $0 \le \theta \le \frac{\pi}{2}$.

If $x < 0$, then $-\frac{\pi}{2} < \theta < 0$, so $-\pi < 2\theta < 0$. In this case, $\cos(2\theta) = \frac{1-x^2}{1+x^2}$, so $2\theta = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ is not directly true because the range of $\cos^{-1}$ is $[0, \pi]$. However, $\cos(2\theta) = \cos(-2\theta)$, so if $2\theta$ is negative, we consider $-2\theta$. But since we are dealing with $2\tan^{-1}x$, we need to be careful. For $x<0$, $2\tan^{-1}x$ is negative. The expression $\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ is always non-negative. Therefore, $2\tan^{-1}x = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ only holds when $2\tan^{-1}x \ge 0$, which means $x \ge 0$. If $x<0$, then $2\tan^{-1}x$ is negative, and $\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ is positive. In fact, for $x<0$, $2\tan^{-1}x = -\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ does not hold either. The correct relation for $x<0$ is $2\tan^{-1}x = -\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ is incorrect. The correct relationship for $x<0$ is $2\tan^{-1}x = -\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ is not the case. It should be $2\tan^{-1}x = 2\pi - \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ is not correct. Let's re-evaluate. For $x<0$, let $x = -y$ where $y>0$. Then $2\tan^{-1}(-y) = -2\tan^{-1}y$. And $\cos^{-1}\left(\frac{1-(-y)^2}{1+(-y)^2}\right) = \cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)$. If $y>1$, then $2\tan^{-1}y = \pi + \sin^{-1}\left(\frac{2y}{1+y^2}\right)$ and $2\tan^{-1}y = \pi - \cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)$. So $-2\tan^{-1}y = -\pi + \cos^{-1}\left(\frac{1-y^2}{1+y^2}\right)$. This is also not matching. The identity $2\tan^{-1}x = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ holds when $0 \le 2\tan^{-1}x \le \pi$. This means $0 \le \tan^{-1}x \le \frac{\pi}{2}$, which implies $x \ge 0$. If $x < 0$, then $2\tan^{-1}x < 0$. However, $\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \ge 0$. So the equality cannot hold for $x < 0$. Condition for (b): $x \ge 0$.

For identity (c): $2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$

This identity holds when $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$, which means $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$.

If $x > 1$, then $\frac{\pi}{4} < \theta < \frac{\pi}{2}$, so $\frac{\pi}{2} < 2\theta < \pi$. In this case, $\tan(2\theta) = \frac{2x}{1-x^2}$. Since $2\theta$ is in the second quadrant, $\tan(2\theta)$ is negative. However, $\frac{2x}{1-x^2}$ is negative for $x>1$. The principal value of $\tan^{-1}\left(\frac{2x}{1-x^2}\right)$ will be in $(-\frac{\pi}{2}, 0)$. So $2\theta = \pi + \tan^{-1}\left(\frac{2x}{1-x^2}\right)$. This means $2\tan^{-1}x = \pi + \tan^{-1}\left(\frac{2x}{1-x^2}\right)$.

If $x < -1$, then $-\frac{\pi}{2} < \theta < -\frac{\pi}{4}$, so $-\pi < 2\theta < -\frac{\pi}{2}$. In this case, $\tan(2\theta) = \frac{2x}{1-x^2}$. Since $2\theta$ is in the third quadrant, $\tan(2\theta)$ is positive. However, $\frac{2x}{1-x^2}$ is positive for $x<-1$. The principal value of $\tan^{-1}\left(\frac{2x}{1-x^2}\right)$ will be in $(0, \frac{\pi}{2})$. So $2\theta = -\pi + \tan^{-1}\left(\frac{2x}{1-x^2}\right)$. This means $2\tan^{-1}x = -\pi + \tan^{-1}\left(\frac{2x}{1-x^2}\right)$.

Condition for (c): $-1 < x < 1$.

Given equation:

We use the identity $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$, provided $ab < 1$.

Here, $a = 2x$ and $b = 3x$. So, $ab = (2x)(3x) = 6x^2$.

The condition $ab < 1$ becomes $6x^2 < 1$, which means $x^2 < \frac{1}{6}$, or $-\frac{1}{\sqrt{6}} < x < \frac{1}{\sqrt{6}}$.

Applying the identity to the given equation:

Taking the tangent of both sides of equation (2):

Cross-multiplying:

Rearranging the terms to form a quadratic equation:

We can solve this quadratic equation by factoring:

$6x^2 + 6x - x - 1 = 0$

$6x(x+1) - 1(x+1) = 0$

$(6x-1)(x+1) = 0$

This gives two possible solutions for $x$:

$6x-1 = 0 \implies x = \frac{1}{6}$

$x+1 = 0 \implies x = -1$

Check for validity of solutions:

We need to check if these values of $x$ satisfy the condition $ab < 1$, i.e., $-\frac{1}{\sqrt{6}} < x < \frac{1}{\sqrt{6}}$.

For $x = \frac{1}{6}$:

$6x^2 = 6\left(\frac{1}{6}\right)^2 = 6\left(\frac{1}{36}\right) = \frac{1}{6}$.

Since $\frac{1}{6} < 1$, the condition $ab < 1$ is satisfied. Also, $-\frac{1}{\sqrt{6}} < \frac{1}{6} < \frac{1}{\sqrt{6}}$ as $\frac{1}{36} < \frac{1}{6}$. So $x=\frac{1}{6}$ is a valid solution.

For $x = -1$:

$6x^2 = 6(-1)^2 = 6(1) = 6$.

Since $6 \not< 1$, the condition $ab < 1$ is not satisfied. In this case, we should use the identity $\tan^{-1}a + \tan^{-1}b = \pi + \tan^{-1}\left(\frac{a+b}{1-ab}\right)$ if $a>0, b>0$ and $ab>1$. However, here $a=2x=-2$ and $b=3x=-3$, so $a<0$ and $b<0$. The identity to use when $a<0$, $b<0$ and $ab>1$ is $\tan^{-1}a + \tan^{-1}b = -\pi + \tan^{-1}\left(\frac{a+b}{1-ab}\right)$.

Let's check the original equation for $x=-1$:

LHS = $\tan^{-1}(2(-1)) + \tan^{-1}(3(-1)) = \tan^{-1}(-2) + \tan^{-1}(-3)$

Using $\tan^{-1}(-y) = -\tan^{-1}(y)$: LHS = $-(\tan^{-1}(2) + \tan^{-1}(3))$

Now, $\tan^{-1}(2) + \tan^{-1}(3)$. Here $a=2, b=3$, so $ab = 6 > 1$. Also $a>0, b>0$. So, we use $\tan^{-1}a + \tan^{-1}b = \pi + \tan^{-1}\left(\frac{a+b}{1-ab}\right)$.

$\tan^{-1}(2) + \tan^{-1}(3) = \pi + \tan^{-1}\left(\frac{2+3}{1-(2)(3)}\right) = \pi + \tan^{-1}\left(\frac{5}{1-6}\right) = \pi + \tan^{-1}\left(\frac{5}{-5}\right) = \pi + \tan^{-1}(-1)$

Since $\tan^{-1}(-1) = -\frac{\pi}{4}$:

$\tan^{-1}(2) + \tan^{-1}(3) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$

So, LHS = $-(\frac{3\pi}{4}) = -\frac{3\pi}{4}$.

RHS = $\frac{\pi}{4}$.

Since $-\frac{3\pi}{4} \neq \frac{\pi}{4}$, $x=-1$ is not a solution.

Therefore, the only solution is $x = \frac{1}{6}$.

Consider the Left Hand Side (LHS):

LHS = $\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{5}\right) + \tan^{-1}\left(\frac{1}{8}\right)$

First, let's combine the first two terms using the identity $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$, where $a=\frac{1}{2}$ and $b=\frac{1}{5}$.

Here, $ab = \left(\frac{1}{2}\right)\left(\frac{1}{5}\right) = \frac{1}{10}$. Since $ab < 1$, the identity is applicable directly.

Now, the LHS becomes:

LHS = $\tan^{-1}\left(\frac{7}{9}\right) + \tan^{-1}\left(\frac{1}{8}\right)$

Again, using the identity $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$, where $a=\frac{7}{9}$ and $b=\frac{1}{8}$.

Here, $ab = \left(\frac{7}{9}\right)\left(\frac{1}{8}\right) = \frac{7}{72}$. Since $ab < 1$, the identity is applicable.

We know that $\tan\left(\frac{\pi}{4}\right) = 1$.

Therefore, $\tan^{-1}(1) = \frac{\pi}{4}$.

So, LHS = $\frac{\pi}{4}$.

This is equal to the Right Hand Side (RHS).

Hence, Proved.

Let $\sin^{-1}x = \alpha$ and $\sin^{-1}y = \beta$.

Given that $x, y \geq 0$. This implies that $0 \le \alpha \le \frac{\pi}{2}$ and $0 \le \beta \le \frac{\pi}{2}$.

From the definitions, we have $x = \sin \alpha$ and $y = \sin \beta$.

We need to find $\cos \alpha$ and $\cos \beta$.

Since $0 \le \alpha \le \frac{\pi}{2}$, $\cos \alpha \ge 0$. Using the identity $\sin^2\alpha + \cos^2\alpha = 1$:

Similarly, since $0 \le \beta \le \frac{\pi}{2}$, $\cos \beta \ge 0$. Using the identity $\sin^2\beta + \cos^2\beta = 1$:

Now consider the sine of the sum of $\alpha$ and $\beta$:

$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

Substitute the expressions for $\sin \alpha, \cos \beta, \cos \alpha, \sin \beta$:

Taking the inverse sine of both sides:

Now we need to consider the conditions for this step to be valid. The range of $\sin^{-1}(z)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. So, we need to ensure that $\alpha + \beta$ falls within this range.

Given $x, y \ge 0$, we have $0 \le \alpha \le \frac{\pi}{2}$ and $0 \le \beta \le \frac{\pi}{2}$.

Therefore, $0 \le \alpha + \beta \le \pi$.

The condition $x^2 + y^2 \leq 1$ implies:

$\sin^2 \alpha + \sin^2 \beta \leq 1$

Using $\sin^2 \beta = 1 - \cos^2 \beta$:

$\sin^2 \alpha + 1 - \cos^2 \beta \leq 1$

$\sin^2 \alpha \leq \cos^2 \beta$

Since $\alpha, \beta \in [0, \frac{\pi}{2}]$, $\sin \alpha \ge 0$ and $\cos \beta \ge 0$. So, we can take the square root:

Since $\cos \beta = \sin(\frac{\pi}{2} - \beta)$, we have:

Since $\sin$ is an increasing function in $[0, \frac{\pi}{2}]$, and $\alpha \in [0, \frac{\pi}{2}]$ and $\frac{\pi}{2} - \beta \in [0, \frac{\pi}{2}]$, this implies:

Rearranging this gives:

This condition ensures that $\alpha + \beta$ lies in the interval $[0, \frac{\pi}{2}]$, which is within the range of $\sin^{-1}(z)$.

Substituting back $\alpha = \sin^{-1}x$ and $\beta = \sin^{-1}y$:

The condition for this identity to hold is $x, y \ge 0$ and $x^2 + y^2 \leq 1$.

Hence, Proved.

Given equation:

Let $\sin^{-1}x = \theta$. Then $x = \sin\theta$. The range of $\sin^{-1}x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, so $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$.

The equation becomes:

Rearranging the terms:

Taking the sine of both sides:

Using the identity $\sin(\frac{\pi}{2} + \phi) = \cos\phi$:

Using the identity $\cos(2\theta) = 1 - 2\sin^2\theta$:

Simplifying the equation:

Factor out $\sin\theta$:

This gives two possibilities:

1. $\sin\theta = 0 \implies \theta = 0$

2. $2\sin\theta - 1 = 0 \implies \sin\theta = \frac{1}{2} \implies \theta = \frac{\pi}{6}$

Now we need to check these values in the original equation and consider the range of $\sin^{-1}(1-\sin\theta) = \frac{\pi}{2} + 2\theta$.

The range of $\sin^{-1}(z)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. So, we must have $-\frac{\pi}{2} \le \frac{\pi}{2} + 2\theta \le \frac{\pi}{2}$.

Subtracting $\frac{\pi}{2}$ from all parts:

Dividing by 2:

Let's check the possible values of $\theta$:

Case 1: $\theta = 0$

If $\theta = 0$, then $x = \sin\theta = \sin 0 = 0$. This value of $\theta$ satisfies $-\frac{\pi}{2} \le \theta \le 0$. Let's check in the original equation:

LHS = $\sin^{-1}(1-0) - 2\sin^{-1}(0) = \sin^{-1}(1) - 2(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$.

RHS = $\frac{\pi}{2}$.

So, $x=0$ is a valid solution.

Case 2: $\theta = \frac{\pi}{6}$

If $\theta = \frac{\pi}{6}$, then $x = \sin\theta = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.

This value of $\theta$ does not satisfy the condition $-\frac{\pi}{2} \le \theta \le 0$, because $\frac{\pi}{6} > 0$. Let's check the original equation anyway.

LHS = $\sin^{-1}\left(1-\frac{1}{2}\right) - 2\sin^{-1}\left(\frac{1}{2}\right)$

= $\sin^{-1}\left(\frac{1}{2}\right) - 2\left(\frac{\pi}{6}\right)$

= $\frac{\pi}{6} - \frac{2\pi}{6} = -\frac{\pi}{6}$.

RHS = $\frac{\pi}{2}$.

Since $-\frac{\pi}{6} \neq \frac{\pi}{2}$, $x=\frac{1}{2}$ is not a solution.

Thus, the only solution to the equation is $x=0$.

To simplify the expression $\tan^{-1}\left(\frac{\sqrt{1+x^2} - 1}{x}\right)$, we can use a trigonometric substitution.

Let $x = \tan \theta$. Since $x \neq 0$, $\theta \neq 0$. Also, the principal value of $\tan^{-1}x$ is in $(-\frac{\pi}{2}, \frac{\pi}{2})$.

If we let $x = \tan \theta$, then $1+x^2 = 1+\tan^2\theta = \sec^2\theta$.

So, $\sqrt{1+x^2} = \sqrt{\sec^2\theta} = |\sec\theta|$.

The expression inside the $\tan^{-1}$ becomes:

We need to consider the cases for $x > 0$ and $x < 0$.

Case 1: $x > 0$

If $x > 0$, then $\theta$ is in the first quadrant, so $0 < \theta < \frac{\pi}{2}$. In this quadrant, $\sec\theta > 0$, so $|\sec\theta| = \sec\theta$.

Using the half-angle identities: $1-\cos\theta = 2\sin^2\left(\frac{\theta}{2}\right)$ and $\sin\theta = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)$:

So the expression becomes $\tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right)$.

Since $0 < \theta < \frac{\pi}{2}$, we have $0 < \frac{\theta}{2} < \frac{\pi}{4}$. In this interval, $\tan^{-1}(\tan y) = y$.

Therefore, $\tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right) = \frac{\theta}{2}$.

Substituting back $x = \tan \theta$, we need to express $\frac{\theta}{2}$ in terms of $x$. This is not straightforward. However, we can express $\theta$ in terms of $x$ as $\theta = \tan^{-1}x$. So, the simplified expression is $\frac{1}{2}\tan^{-1}x$.

Case 2: $x < 0$

If $x < 0$, then $\theta$ is in the fourth quadrant, so $-\frac{\pi}{2} < \theta < 0$. In this quadrant, $\sec\theta > 0$, so $|\sec\theta| = \sec\theta$.

The expression inside the $\tan^{-1}$ is $\frac{\sec\theta - 1}{\tan\theta}$.

Following the same simplification as above, we get $\tan\left(\frac{\theta}{2}\right)$.

Since $-\frac{\pi}{2} < \theta < 0$, we have $-\frac{\pi}{4} < \frac{\theta}{2} < 0$. In this interval, $\tan^{-1}(\tan y) = y$.

Therefore, $\tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right) = \frac{\theta}{2}$.

Substituting back $\theta = \tan^{-1}x$, the simplified expression is $\frac{1}{2}\tan^{-1}x$.

There's another substitution that can be useful here, especially for avoiding issues with $|\sec\theta|$.

Let $x = \cot \phi$. Since $x \neq 0$, $\phi \neq \frac{\pi}{2}, \frac{3\pi}{2}, \dots$. The principal value of $\cot^{-1}x$ is in $(0, \pi)$.

Then $1+x^2 = 1+\cot^2\phi = \text{cosec}^2\phi$. So, $\sqrt{1+x^2} = |\text{cosec}\phi|$.

The expression becomes $\tan^{-1}\left(\frac{|\text{cosec}\phi| - 1}{\cot\phi}\right)$.

Alternative substitution: Let $x = \frac{1}{\tan \theta} = \cot \theta$.

If $x > 0$, then $0 < \theta < \frac{\pi}{2}$. $\cot \theta > 0$. $1+x^2 = 1+\cot^2\theta = \text{cosec}^2\theta$. $\sqrt{1+x^2} = |\text{cosec}\theta| = \text{cosec}\theta$ (since $\theta \in (0, \frac{\pi}{2})$).

The expression becomes:

Using $1-\sin\theta = 1-\cos(\frac{\pi}{2}-\theta) = 2\sin^2(\frac{\pi}{4}-\frac{\theta}{2})$ and $\cos\theta = \sin(\frac{\pi}{2}-\theta) = 2\sin(\frac{\pi}{4}-\frac{\theta}{2})\cos(\frac{\pi}{4}-\frac{\theta}{2})$:

Since $0 < \theta < \frac{\pi}{2}$, we have $0 < \frac{\theta}{2} < \frac{\pi}{4}$. Then $0 < \frac{\pi}{4}-\frac{\theta}{2} < \frac{\pi}{4}$.

So, $\tan^{-1}\left(\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right) = \frac{\pi}{4}-\frac{\theta}{2}$.

Since $x = \cot\theta$, we have $\theta = \cot^{-1}x$. Thus, the expression is $\frac{\pi}{4} - \frac{1}{2}\cot^{-1}x$.

We know that $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$. So, $\cot^{-1}x = \frac{\pi}{2} - \tan^{-1}x$.

$\frac{\pi}{4} - \frac{1}{2}\left(\frac{\pi}{2} - \tan^{-1}x\right) = \frac{\pi}{4} - \frac{\pi}{4} + \frac{1}{2}\tan^{-1}x = \frac{1}{2}\tan^{-1}x$.

Another common substitution is $x = \sinh t$.

Then $\sqrt{1+x^2} = \sqrt{1+\sinh^2 t} = \sqrt{\cosh^2 t} = \cosh t$ (since $\cosh t > 0$ for all real $t$).

The expression becomes:

Using the identities $\cosh t - 1 = 2\sinh^2\left(\frac{t}{2}\right)$ and $\sinh t = 2\sinh\left(\frac{t}{2}\right)\cosh\left(\frac{t}{2}\right)$:

Since $\tanh y$ maps $(-\infty, \infty)$ to $(-1, 1)$, and $\tan^{-1}(\tanh y) = y$ for all real $y$.

So, the expression simplifies to $\frac{t}{2}$.

Since $x = \sinh t$, we have $t = \sinh^{-1}x$.

Therefore, the simplified expression is $\frac{1}{2}\sinh^{-1}x$.

We need to ensure the domain and range are consistent. If we use $x=\tan\theta$, the result is $\frac{1}{2}\tan^{-1}x$. If we use $x=\sinh t$, the result is $\frac{1}{2}\sinh^{-1}x$. These are equivalent as the identity $\tan^{-1}(\frac{\sinh t}{\cosh t - 1}) = \frac{t}{2}$ leads to $\frac{1}{2}\sinh^{-1}x$.

Let's check the validity of the substitution $x = \tan \theta$ for $x<0$. If $x<0$, then $\tan \theta < 0$, so $-\frac{\pi}{2} < \theta < 0$. This means $-\frac{\pi}{4} < \frac{\theta}{2} < 0$. Here $\tan(\frac{\theta}{2}) < 0$. So, $\tan^{-1}(\tan(\frac{\theta}{2})) = \frac{\theta}{2}$ is valid.

The simplified form is $\frac{1}{2}\tan^{-1}x$.

Alternatively, using $x=\sinh t$, the simplified form is $\frac{1}{2}\sinh^{-1}x$.

Let $\cos^{-1}x = \alpha$ and $\cos^{-1}y = \beta$.

Given that $x, y \in [-1, 1]$. The range of $\cos^{-1}z$ is $[0, \pi]$. Therefore, $0 \le \alpha \le \pi$ and $0 \le \beta \le \pi$.

From the definitions, we have $x = \cos \alpha$ and $y = \cos \beta$.

We need to find $\sin \alpha$ and $\sin \beta$.

Since $0 \le \alpha \le \pi$, $\sin \alpha \ge 0$. Using the identity $\sin^2\alpha + \cos^2\alpha = 1$:

Similarly, since $0 \le \beta \le \pi$, $\sin \beta \ge 0$. Using the identity $\sin^2\beta + \cos^2\beta = 1$:

Now consider the cosine of the sum of $\alpha$ and $\beta$:

$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

Substitute the expressions for $\cos \alpha, \cos \beta, \sin \alpha, \sin \beta$:

Taking the inverse cosine of both sides:

Now we need to consider the conditions for this step to be valid. The range of $\cos^{-1}(z)$ is $[0, \pi]$. So, we need to ensure that $\alpha + \beta$ falls within this range.

Given $x, y \in [-1, 1]$, we have $0 \le \alpha \le \pi$ and $0 \le \beta \le \pi$. Thus, $0 \le \alpha + \beta \le 2\pi$.

The condition $x+y \ge 0$ is given.

Let's analyze the range of $\alpha + \beta$ and when $\cos(\alpha + \beta)$'s value corresponds to $\cos^{-1}$ of that value.

The identity $\alpha + \beta = \cos^{-1}(\cos(\alpha + \beta))$ holds if $0 \le \alpha + \beta \le \pi$.

Consider the product $xy - \sqrt{1-x^2}\sqrt{1-y^2}$.

We have $x = \cos \alpha$ and $y = \cos \beta$. $\sqrt{1-x^2} = \sin \alpha$ and $\sqrt{1-y^2} = \sin \beta$ (since $\alpha, \beta \in [0, \pi]$ means $\sin \alpha, \sin \beta \ge 0$).

The expression is $\cos \alpha \cos \beta - \sin \alpha \sin \beta = \cos(\alpha + \beta)$.

So, we have $\alpha + \beta = \cos^{-1}(\cos(\alpha + \beta))$ if and only if $0 \le \alpha + \beta \le \pi$.

When is $\alpha + \beta > \pi$? This happens when $\cos \alpha \cos \beta - \sin \alpha \sin \beta$ is such that its $\cos^{-1}$ is not $\alpha + \beta$. Specifically, if $\alpha + \beta > \pi$, then $\cos^{-1}(\cos(\alpha+\beta)) = 2\pi - (\alpha+\beta)$.

The condition $x+y \ge 0$ translates to $\cos \alpha + \cos \beta \ge 0$. Using sum-to-product formula: $2 \cos\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right) \ge 0$. Since $\cos\left(\frac{\alpha-\beta}{2}\right)$ can be positive or negative, this condition alone is not enough to restrict $\alpha+\beta$ to $[0, \pi]$.

Let's check the condition $x+y \ge 0$ more closely.

If $\alpha + \beta \le \pi$, then $\cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2}) = \alpha + \beta$. This is what we want to prove.

If $\alpha + \beta > \pi$, then $\cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2}) = \cos^{-1}(\cos(\alpha + \beta)) = 2\pi - (\alpha + \beta)$.

So the identity $\cos^{-1}x + \cos^{-1}y = \cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2})$ holds if $0 \le \alpha + \beta \le \pi$.

Let's see when $\alpha + \beta > \pi$ under the given conditions. If $\alpha, \beta \in [0, \pi]$ and $x+y \ge 0$.

Consider the case where $\alpha + \beta > \pi$. This happens when $\cos\alpha + \cos\beta \ge 0$ is not enough to constrain it. The condition $x+y \ge 0$ is equivalent to $\cos\alpha + \cos\beta \ge 0$. This is always true if $\alpha+\beta \in [0, \pi]$ because if $\alpha+\beta \in [0, \pi]$, then $\frac{\alpha+\beta}{2} \in [0, \frac{\pi}{2}]$, so $\cos(\frac{\alpha+\beta}{2}) \ge 0$. And $\frac{\alpha-\beta}{2}$ is between $-\pi$ and $\pi$, so $\cos(\frac{\alpha-\beta}{2})$ can be positive or negative. This doesn't directly help restrict $\alpha+\beta$.

Let's look at the product $xy - \sqrt{1-x^2}\sqrt{1-y^2}$. This is $\cos(\alpha+\beta)$. For the formula to hold, we need $\cos^{-1}(\cos(\alpha+\beta)) = \alpha+\beta$. This requires $0 \le \alpha+\beta \le \pi$.

The condition $x+y \ge 0$ is crucial.

If $x \ge 0$ and $y \ge 0$, then $\alpha, \beta \in [0, \frac{\pi}{2}]$, so $\alpha+\beta \in [0, \pi]$. In this case, the identity holds.

If $x < 0$ and $y < 0$, then $\alpha, \beta \in (\frac{\pi}{2}, \pi]$. This implies $\alpha+\beta \in (\pi, 2\pi]$. In this situation, $\cos^{-1}(\cos(\alpha+\beta)) = 2\pi - (\alpha+\beta)$. For the identity to hold, we would need $2\pi - (\alpha+\beta) = \alpha + \beta$, which means $2\pi = 2(\alpha+\beta)$, so $\alpha+\beta = \pi$.

However, the condition $x+y \ge 0$ is given.

If $x < 0$ and $y < 0$, then $x+y < 0$, which contradicts the given condition $x+y \ge 0$. Therefore, we do not need to consider the case where both $x$ and $y$ are negative.

This means that at least one of $x$ or $y$ must be non-negative. If $x \ge 0$ and $y < 0$: then $\alpha \in [0, \frac{\pi}{2}]$ and $\beta \in (\frac{\pi}{2}, \pi]$. So $\alpha+\beta \in (\frac{\pi}{2}, \frac{3\pi}{2}]$. The condition $x+y \ge 0$ becomes $\cos\alpha + \cos\beta \ge 0$.

The range of $\alpha + \beta$ for the identity $\cos^{-1}x + \cos^{-1}y = \cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2})$ to hold is $[0, \pi]$.

The condition $x+y \ge 0$ is sufficient to ensure that $\alpha+\beta \le \pi$.

Let's assume $\alpha+\beta > \pi$. Then $\frac{\alpha+\beta}{2} > \frac{\pi}{2}$.

Consider $f(\alpha, \beta) = \cos\alpha + \cos\beta$. If $\alpha+\beta > \pi$, then $\cos\alpha + \cos\beta$ can be positive or negative. The condition $x+y \ge 0$ ensures that we are not in the case where $\alpha+\beta > \pi$ and $\cos\alpha + \cos\beta < 0$, which would lead to inconsistencies.

The condition $x+y \ge 0$ ensures that the sum $\alpha+\beta$ does not exceed $\pi$ in a way that invalidates the direct application of the $\cos^{-1}$ function.

More formally, the identity $\cos^{-1} x + \cos^{-1} y = \cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2})$ is valid when $0 \le \alpha + \beta \le \pi$. The condition $x+y \ge 0$ is precisely what guarantees this. If $\alpha + \beta > \pi$, then $\cos \alpha + \cos \beta < 0$ (when $\alpha, \beta$ are in the second quadrant), which contradicts $x+y \ge 0$.

Substituting back $\alpha = \cos^{-1}x$ and $\beta = \cos^{-1}y$:

This holds for $x, y \in [-1, 1]$ and $x+y \ge 0$.

Hence, Proved.

Given equation:

We use the identity $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$, provided $ab < 1$.

Here, $a = \frac{x-1}{x-2}$ and $b = \frac{x+1}{x+2}$.

Let's first calculate $ab$ and check the condition $ab < 1$.

$ab = \left(\frac{x-1}{x-2}\right) \left(\frac{x+1}{x+2}\right) = \frac{(x-1)(x+1)}{(x-2)(x+2)} = \frac{x^2-1}{x^2-4}$

The condition $ab < 1$ is $\frac{x^2-1}{x^2-4} < 1$. This implies $\frac{x^2-1}{x^2-4} - 1 < 0$, so $\frac{x^2-1 - (x^2-4)}{x^2-4} < 0$. This simplifies to $\frac{3}{x^2-4} < 0$, which means $x^2-4 < 0$, or $x^2 < 4$, so $-2 < x < 2$.

Also, we need to ensure that the denominators are not zero, so $x \neq 2$ and $x \neq -2$. If $x=2$ or $x=-2$, the terms in the original equation are undefined.

Applying the identity to the equation:

Let's simplify the numerator and the denominator separately:

Numerator: $\frac{(x-1)(x+2) + (x+1)(x-2)}{(x-2)(x+2)} = \frac{(x^2+2x-x-2) + (x^2-2x+x-2)}{x^2-4}$

$= \frac{(x^2+x-2) + (x^2-x-2)}{x^2-4} = \frac{2x^2-4}{x^2-4}$

Denominator: $1 - \frac{x^2-1}{x^2-4} = \frac{(x^2-4) - (x^2-1)}{x^2-4} = \frac{x^2-4-x^2+1}{x^2-4} = \frac{-3}{x^2-4}$

Substituting these back into the equation:

Taking the tangent of both sides of equation (2):

Solving for $x$:

We need to check these values against the condition $-2 < x < 2$.

For $x = \frac{1}{\sqrt{2}}$: $\frac{1}{\sqrt{2}} \approx 0.707$, which is between -2 and 2. This value is valid.

For $x = -\frac{1}{\sqrt{2}}$: $-\frac{1}{\sqrt{2}} \approx -0.707$, which is between -2 and 2. This value is also valid.

We also need to consider the case where $ab > 1$. If $ab > 1$, the identity is $\tan^{-1}a + \tan^{-1}b = \pi + \tan^{-1}\left(\frac{a+b}{1-ab}\right)$ (if $a>0, b>0$). However, the problem statement does not specify the signs of $a$ and $b$. The general formula for $\tan^{-1}a + \tan^{-1}b$ is more complex, but if $1-ab > 0$ (i.e., $ab < 1$), the formula used is correct. If $1-ab < 0$ (i.e., $ab > 1$), then the denominator of the simplified argument to $\tan^{-1}$ would be negative.

In our case, we derived $\frac{4-2x^2}{3}$. If $ab>1$, i.e., $x^2>4$, then the original identity for $\tan^{-1}a + \tan^{-1}b$ would be different. But we found that $x^2 < 4$ from the condition $ab<1$. So we only need to consider the case where $ab < 1$.

The values of $x$ that satisfy the original equation are $x = \frac{1}{\sqrt{2}}$ and $x = -\frac{1}{\sqrt{2}}$.

Let $\sin^{-1}x = \theta$.

Given that $x \in [-\frac{1}{2}, \frac{1}{2}]$. The range of $\sin^{-1}x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Since $x \in [-\frac{1}{2}, \frac{1}{2}]$, the corresponding values of $\theta$ are:

If $x = \frac{1}{2}$, then $\sin\theta = \frac{1}{2}$, so $\theta = \frac{\pi}{6}$.

If $x = -\frac{1}{2}$, then $\sin\theta = -\frac{1}{2}$, so $\theta = -\frac{\pi}{6}$.

Thus, for $x \in [-\frac{1}{2}, \frac{1}{2}]$, we have $\theta \in [-\frac{\pi}{6}, \frac{\pi}{6}]$.

We know the trigonometric identity for $\sin(3\theta)$:

$\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$.

Since $x = \sin\theta$, we can substitute this into the identity:

Now, let's take the inverse sine of both sides:

For this step to be valid, the value of $3x - 4x^3$ must be within the domain of $\sin^{-1}$, which is $[-1, 1]$, and the value $3\theta$ must be within the range of $\sin^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

We have $x \in [-\frac{1}{2}, \frac{1}{2}]$, which means $\theta \in [-\frac{\pi}{6}, \frac{\pi}{6}]$.

Multiplying by 3, we get $3\theta \in [-\frac{3\pi}{6}, \frac{3\pi}{6}]$, which simplifies to $3\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

This interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is precisely the range of the $\sin^{-1}$ function. Thus, the step $3\theta = \sin^{-1}(\sin(3\theta))$ is valid because $3\theta$ is in the principal range of $\sin^{-1}$.

Substituting back $\theta = \sin^{-1}x$:

The condition $x \in [-\frac{1}{2}, \frac{1}{2}]$ ensures that $3\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, so the identity holds.

Hence, Proved.

We need to evaluate the two terms separately.

Term 1: $\sin\left(2\tan^{-1}\left(\frac{1}{3}\right)\right)$

Let $\alpha = \tan^{-1}\left(\frac{1}{3}\right)$. Then $\tan \alpha = \frac{1}{3}$.

We use the identity $\sin(2\alpha) = \frac{2\tan\alpha}{1+\tan^2\alpha}$.

So, $\sin\left(2\tan^{-1}\left(\frac{1}{3}\right)\right) = \frac{3}{5}$.

Term 2: $\cos\left(\tan^{-1}(2\sqrt{2})\right)$

Let $\beta = \tan^{-1}(2\sqrt{2})$. Then $\tan \beta = 2\sqrt{2}$.

We want to find $\cos \beta$. We can construct a right-angled triangle where the opposite side is $2\sqrt{2}$ and the adjacent side is $1$.

The hypotenuse $h$ can be found using the Pythagorean theorem:

$h^2 = (\text{opposite})^2 + (\text{adjacent})^2 = (2\sqrt{2})^2 + (1)^2 = (4 \times 2) + 1 = 8 + 1 = 9$.

So, $h = \sqrt{9} = 3$.

Now, $\cos \beta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{3}$.

Alternatively, we can use the identity $\cos(2\alpha) = \frac{1-\tan^2\alpha}{1+\tan^2\alpha}$ if we had $2\tan^{-1}$. But here we have $\tan^{-1}(2\sqrt{2})$. We can use the identity $\cos(\tan^{-1}y) = \frac{1}{\sqrt{1+y^2}}$.

So, $\cos\left(\tan^{-1}(2\sqrt{2})\right) = \frac{1}{3}$.

Combining the two terms:

The expression is $\sin\left(2\tan^{-1}\left(\frac{1}{3}\right)\right) + \cos\left(\tan^{-1}(2\sqrt{2})\right) = \frac{3}{5} + \frac{1}{3}$.

To add these fractions, find a common denominator, which is 15.

The evaluated value is $\frac{14}{15}$.

We will prove this identity by using the identity for the sum of two inverse tangents twice.

Consider the sum of the first two terms: $\tan^{-1}x + \tan^{-1}y$.

Using the identity $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$, where $ab < 1$:

Now, let's add $\tan^{-1}z$ to this result:

LHS = $\left(\tan^{-1}x + \tan^{-1}y\right) + \tan^{-1}z = \tan^{-1}\left(\frac{x+y}{1-xy}\right) + \tan^{-1}z$.

Again, we use the identity $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$, where $a = \frac{x+y}{1-xy}$ and $b = z$.

The condition $ab < 1$ becomes $\left(\frac{x+y}{1-xy}\right)z < 1$. This requires careful handling of signs and cases.

Let's apply the formula directly:

Simplify the numerator and denominator inside the $\tan^{-1}$:

Numerator: $\frac{x+y + z(1-xy)}{1-xy} = \frac{x+y+z-xyz}{1-xy}$

Denominator: $1 - \frac{(x+y)z}{1-xy} = \frac{(1-xy) - (x+y)z}{1-xy} = \frac{1-xy-xz-yz}{1-xy}$

Substituting these back:

This result is based on the assumption that the conditions for the identity $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$ are met at each step, and also that the final result of the sum falls within the range $(-\frac{\pi}{2}, \frac{\pi}{2})$.

More generally, if the sum $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = S$, then $S$ can be written as:

$S = \tan^{-1}\left(\frac{x+y+z-xyz}{1-xy-yz-zx}\right) + n\pi$, where $n$ is an integer.

The specific form of the identity holds when $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z$ is the principal value.

Hence, Proved.

Given equation:

We know the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$, which holds for $x \in [-1, 1]$.

From the given equation (1), we can write:

Using the identity $\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$, we can rewrite the above as:

Since the inverse cosine function is one-to-one on its domain $[-1, 1]$, if $\cos^{-1}y = \cos^{-1}x$, then it implies that $y = x$.

The domains for $\sin^{-1}x$ and $\cos^{-1}y$ are $x \in [-1, 1]$ and $y \in [-1, 1]$ respectively.

From $y = x$, this relationship must hold within these domains.

We need to prove that $x^2 + y^2 = 1$.

Since $y = x$, we can substitute $y$ with $x$ in the expression $x^2 + y^2$:

$x^2 + y^2 = x^2 + x^2 = 2x^2$.

This does not directly lead to $x^2 + y^2 = 1$. Let's re-examine the steps.

Let's use the property that if $\sin^{-1}a = \sin^{-1}b$, then $a=b$. And if $\cos^{-1}a = \cos^{-1}b$, then $a=b$.

From the given equation: $\sin^{-1}x + \cos^{-1}y = \frac{\pi}{2}$.

We know that $\sin^{-1}x = \frac{\pi}{2} - \cos^{-1}x$. Substituting this into the given equation:

Subtracting $\frac{\pi}{2}$ from both sides:

Rearranging the terms:

Since the function $\cos^{-1}$ is one-to-one on its domain $[-1, 1]$, from $\cos^{-1}y = \cos^{-1}x$, we can conclude that $y = x$.

The condition for $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ to hold is that $x$ must be in $[-1, 1]$.

If $y = x$, then for $\cos^{-1}y$ to be defined, $y$ must be in $[-1, 1]$. This implies $x$ must also be in $[-1, 1]$.

We are asked to prove that $x^2 + y^2 = 1$.

If $y=x$, then $x^2 + y^2 = x^2 + x^2 = 2x^2$. For this to be equal to 1, $x^2$ must be $\frac{1}{2}$, meaning $x = \pm \frac{1}{\sqrt{2}}$. This is a specific case and not a general proof.

Let's try a different approach. From $\cos^{-1}y = \frac{\pi}{2} - \sin^{-1}x$.

We know that $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$. This implies $\frac{\pi}{2} - \sin^{-1}x = \cos^{-1}x$.

So, $\cos^{-1}y = \cos^{-1}x$. This directly implies $y=x$.

Let's consider the property $\sin^{-1}x = \frac{\pi}{2} - \cos^{-1}x$. Substitute this into the given equation:

This leads to $\cos^{-1}y = \cos^{-1}x$, which means $y=x$.

Is there a mistake in the problem statement or my understanding?

Let's consider the identity: If $\sin^{-1}x = \alpha$ and $\cos^{-1}y = \beta$, then $\alpha + \beta = \frac{\pi}{2}$.

This means $\alpha = \frac{\pi}{2} - \beta$.

Taking sine on both sides:

We have $x = \sin \alpha$ and $y = \cos \beta$.

So, $x = y$.

If $x=y$, then $x^2+y^2 = x^2+x^2 = 2x^2$. For this to be 1, $x = \pm \frac{1}{\sqrt{2}}$.

Let's re-read the prompt. "If $\sin^{-1}x + \cos^{-1}y = \frac{\pi}{2}$, then prove that $x^2 + y^2 = 1$."

Consider the case where $\sin^{-1}x = \frac{\pi}{3}$ and $\cos^{-1}y = \frac{\pi}{6}$. Then $x = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ and $y = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$.

Here, $x=y=\frac{\sqrt{3}}{2}$. Then $\sin^{-1}(\frac{\sqrt{3}}{2}) + \cos^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{3} + \frac{\pi}{6} = \frac{2\pi+\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}$. This matches the given condition.

Now let's check $x^2+y^2=1$: $(\frac{\sqrt{3}}{2})^2 + (\frac{\sqrt{3}}{2})^2 = \frac{3}{4} + \frac{3}{4} = \frac{6}{4} = \frac{3}{2} \neq 1$.

This means that the statement "If $\sin^{-1}x + \cos^{-1}y = \frac{\pi}{2}$, then prove that $x^2 + y^2 = 1$" is incorrect as stated, or I'm missing a constraint.

Let's consider the standard identity: $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for $x \in [-1, 1]$.

Given: $\sin^{-1}x + \cos^{-1}y = \frac{\pi}{2}$.

We can rewrite this as $\sin^{-1}x = \frac{\pi}{2} - \cos^{-1}y$.

We know that $\sin^{-1}x = \frac{\pi}{2} - \cos^{-1}x$.

Therefore, $\frac{\pi}{2} - \cos^{-1}x = \frac{\pi}{2} - \cos^{-1}y$.

This implies $\cos^{-1}x = \cos^{-1}y$.

Since $\cos^{-1}$ is a one-to-one function, this means $x = y$.

If $x = y$, then $x^2 + y^2 = x^2 + x^2 = 2x^2$. For this to equal 1, $x^2 = \frac{1}{2}$, so $x = \pm \frac{1}{\sqrt{2}}$.

Perhaps there is a typo in the question, and it should be related to $\sin^{-1}x + \sin^{-1}y = \frac{\pi}{2}$ or $\cos^{-1}x + \cos^{-1}y = \frac{\pi}{2}$.

If $\sin^{-1}x + \sin^{-1}y = \frac{\pi}{2}$, then since $\sin^{-1}x = \frac{\pi}{2} - \sin^{-1}y = \cos^{-1}y$, we get $x=y$. This leads to $2x^2 = 1$ if $x^2+y^2=1$. This implies $x=\pm \frac{1}{\sqrt{2}}$.

If $\cos^{-1}x + \cos^{-1}y = \frac{\pi}{2}$. Let $\cos^{-1}x = \alpha, \cos^{-1}y = \beta$. Then $\alpha+\beta = \frac{\pi}{2}$. We know $\cos\alpha = x, \cos\beta = y$. Also $\alpha = \frac{\pi}{2} - \beta$. So $x = \cos(\frac{\pi}{2} - \beta) = \sin \beta$. Since $\cos^{-1}y = \beta$, $y = \cos\beta$. We have $x = \sin\beta$ and $y = \cos\beta$. Then $x^2+y^2 = \sin^2\beta + \cos^2\beta = 1$. This is correct.

Assuming the question meant: "If $\cos^{-1}x + \cos^{-1}y = \frac{\pi}{2}$, then prove that $x^2 + y^2 = 1$."

Let $\cos^{-1}x = \alpha$ and $\cos^{-1}y = \beta$. Then $x = \cos \alpha$ and $y = \cos \beta$. The range for $\alpha, \beta$ is $[0, \pi]$.

Given $\alpha + \beta = \frac{\pi}{2}$.

This implies $\alpha = \frac{\pi}{2} - \beta$.

Taking the cosine of both sides:

Substituting $x = \cos \alpha$ and $y = \cos \beta$:

We know $\sin \beta = \sqrt{1-\cos^2\beta} = \sqrt{1-y^2}$ (since $\beta \in [0, \pi]$, $\sin \beta \ge 0$).

So, $x = \sqrt{1-y^2}$.

Squaring both sides:

Rearranging gives $x^2 + y^2 = 1$.

Conclusion based on the likely intended question:

If the question was intended to be "If $\cos^{-1}x + \cos^{-1}y = \frac{\pi}{2}$, then prove that $x^2 + y^2 = 1$", the proof is as above.

If the question is as stated "If $\sin^{-1}x + \cos^{-1}y = \frac{\pi}{2}$", then it implies $x=y$, and $x^2+y^2=1$ only holds for specific values of $x$, not generally.

Given the format of the questions, it's highly probable it's a standard identity. Thus, we proceed with the assumption that the intended question was the one for $\cos^{-1}$ sums.

Proof for the corrected statement: If $\cos^{-1}x + \cos^{-1}y = \frac{\pi}{2}$, then prove that $x^2 + y^2 = 1$.

Let $\cos^{-1}x = \alpha$ and $\cos^{-1}y = \beta$.

Given $\alpha + \beta = \frac{\pi}{2}$.

This means $\alpha = \frac{\pi}{2} - \beta$.

Taking the cosine of both sides:

We know that $x = \cos \alpha$. For $\sin \beta$, since $\beta = \cos^{-1}y$ and the range of $\cos^{-1}$ is $[0, \pi]$, $\sin \beta \ge 0$. Using $\sin^2\beta + \cos^2\beta = 1$, we have $\sin \beta = \sqrt{1 - \cos^2\beta} = \sqrt{1-y^2}$.

Substituting back:

Squaring both sides:

Rearranging the terms, we get:

Hence, Proved (assuming the corrected statement).