Chapter 2 Inverse Trigonometric Functions

To Find:

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

Solution:

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

By definition of the inverse sine function, this implies:

$\sin(y) = \frac{1}{\sqrt{2}}$

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

We need to find the angle $y$ within this interval such that $\sin(y) = \frac{1}{\sqrt{2}}$.

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

Since $\frac{\pi}{4}$ lies in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, it is the principal value.

Therefore, $y = \frac{\pi}{4}$.

The principal value of $\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$ is $\mathbf{\frac{\pi}{4}}$.

To Find:

The principal value of $\cot^{-1}\left(\frac{-1}{\sqrt{3}}\right)$.

Solution:

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

By the definition of the inverse cotangent function, this implies:

$\cot(y) = -\frac{1}{\sqrt{3}}$

We know that the range of the principal value branch of $\cot^{-1}(x)$ is $(0, \pi)$.

We need to find the angle $y$ within the interval $(0, \pi)$ such that $\cot(y) = -\frac{1}{\sqrt{3}}$.

First, let's find the angle $\alpha$ in the first quadrant $(0, \frac{\pi}{2})$ such that $\cot(\alpha) = \frac{1}{\sqrt{3}}$.

$\cot(\alpha) = \frac{1}{\sqrt{3}}$ implies $\tan(\alpha) = \sqrt{3}$.

The angle for which $\tan(\alpha) = \sqrt{3}$ is $\alpha = \frac{\pi}{3}$.

Since $\cot(y)$ is negative $\left(-\frac{1}{\sqrt{3}}\right)$, and the principal value range is $(0, \pi)$, the angle $y$ must lie in the second quadrant (where cotangent is negative).

The angle in the second quadrant with reference angle $\alpha = \frac{\pi}{3}$ is given by:

$y = \pi - \alpha$

$y = \pi - \frac{\pi}{3}$

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

We check if this value lies in the principal value range $(0, \pi)$. Since $0 < \frac{2\pi}{3} < \pi$, it does.

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

The principal value of $\cot^{-1}\left(\frac{-1}{\sqrt{3}}\right)$ is $\mathbf{\frac{2\pi}{3}}$.

To Find:

The principal value of $\sin^{-1}\left(-\frac{1}{2}\right)$.

Solution:

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

By the definition of the inverse sine function, this means:

$\sin(y) = -\frac{1}{2}$

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

We need to find the angle $y$ within this interval such that $\sin(y) = -\frac{1}{2}$.

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

Since $\sin(y)$ is negative and $y$ must be in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, $y$ must be in the fourth quadrant part of this interval, i.e., $y \in \left[-\frac{\pi}{2}, 0\right)$.

The angle in this range whose sine is $-\frac{1}{2}$ is related to $\frac{\pi}{6}$ by $\sin(-x) = -\sin(x)$.

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

The value $y = -\frac{\pi}{6}$ lies in the principal value range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Therefore, the principal value of $\sin^{-1}\left(-\frac{1}{2}\right)$ is $-\frac{\pi}{6}$.

The principal value of $\sin^{-1}\left(-\frac{1}{2}\right)$ is $\mathbf{-\frac{\pi}{6}}$.

To Find:

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

Solution:

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

By the definition of the inverse cosine function, this means:

$\cos(y) = \frac{\sqrt{3}}{2}$

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

We need to find the angle $y$ within this interval such that $\cos(y) = \frac{\sqrt{3}}{2}$.

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

The angle $y = \frac{\pi}{6}$ lies in the principal value range $[0, \pi]$.

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

The principal value of $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$ is $\mathbf{\frac{\pi}{6}}$.

To Find:

The principal value of $\text{cosec}^{-1}(2)$.

Solution:

Let $y = \text{cosec}^{-1}(2)$.

By the definition of the inverse cosecant function, this means:

$\text{cosec}(y) = 2$

Since $\text{cosec}(y) = \frac{1}{\sin(y)}$, we have:

$\frac{1}{\sin(y)} = 2$

$\implies \sin(y) = \frac{1}{2}$

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

We need to find the angle $y$ within this range such that $\sin(y) = \frac{1}{2}$.

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

The angle $y = \frac{\pi}{6}$ lies in the interval $\left(0, \frac{\pi}{2}\right]$, which is part of the principal value range for $\text{cosec}^{-1}$.

Therefore, the principal value of $\text{cosec}^{-1}(2)$ is $\frac{\pi}{6}$.

The principal value of $\text{cosec}^{-1}(2)$ is $\mathbf{\frac{\pi}{6}}$.

To Find:

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

Solution:

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

By the definition of the inverse tangent function, this means:

$\tan(y) = -\sqrt{3}$

The principal value branch of $\tan^{-1}(x)$ has the range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

We need to find the angle $y$ within this interval such that $\tan(y) = -\sqrt{3}$.

First, consider the positive value. We know that $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$.

Since $\tan(y)$ is negative and $y$ must be in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, the angle $y$ must lie in the interval $\left(-\frac{\pi}{2}, 0\right)$.

We use the property $\tan(-x) = -\tan(x)$.

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

The angle $y = -\frac{\pi}{3}$ lies within the principal value range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

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

The principal value of $\tan^{-1}(-\sqrt{3})$ is $\mathbf{-\frac{\pi}{3}}$.

To Find:

The principal value of $\cos^{-1}\left(-\frac{1}{2}\right)$.

Solution:

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

By the definition of the inverse cosine function, this means:

$\cos(y) = -\frac{1}{2}$

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

We need to find the angle $y$ within this interval such that $\cos(y) = -\frac{1}{2}$.

First, let's find the reference angle $\alpha$ in the first quadrant $[0, \frac{\pi}{2}]$ such that $\cos(\alpha) = \frac{1}{2}$. We know that $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$. So, the reference angle is $\alpha = \frac{\pi}{3}$.

Since $\cos(y)$ is negative ($-\frac{1}{2}$) and the principal value range is $[0, \pi]$, the angle $y$ must lie in the second quadrant ($\frac{\pi}{2} < y \le \pi$), where cosine is negative.

The angle in the second quadrant with reference angle $\alpha = \frac{\pi}{3}$ is given by:

$y = \pi - \alpha$

$y = \pi - \frac{\pi}{3}$

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

The angle $y = \frac{2\pi}{3}$ lies within the principal value range $[0, \pi]$.

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

The principal value of $\cos^{-1}\left(-\frac{1}{2}\right)$ is $\mathbf{\frac{2\pi}{3}}$.

To Find:

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

Solution:

Let $y = \tan^{-1}(-1)$.

By the definition of the inverse tangent function, this means:

$\tan(y) = -1$

The principal value branch of $\tan^{-1}(x)$ has the range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

We need to find the angle $y$ within this interval such that $\tan(y) = -1$.

First, consider the positive value. We know that $\tan\left(\frac{\pi}{4}\right) = 1$.

Since $\tan(y)$ is negative and $y$ must be in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, the angle $y$ must lie in the interval $\left(-\frac{\pi}{2}, 0\right)$.

We use the property $\tan(-x) = -\tan(x)$.

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

The angle $y = -\frac{\pi}{4}$ lies within the principal value range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

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

The principal value of $\tan^{-1}(-1)$ is $\mathbf{-\frac{\pi}{4}}$.

To Find:

The principal value of $\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)$.

Solution:

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

By the definition of the inverse secant function, this means:

$\sec(y) = \frac{2}{\sqrt{3}}$

Since $\sec(y) = \frac{1}{\cos(y)}$, we have:

$\frac{1}{\cos(y)} = \frac{2}{\sqrt{3}}$

$\implies \cos(y) = \frac{\sqrt{3}}{2}$

The principal value branch of $\sec^{-1}(x)$ has the range $\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]$.

We need to find the angle $y$ within this range such that $\cos(y) = \frac{\sqrt{3}}{2}$.

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

The angle $y = \frac{\pi}{6}$ lies in the interval $\left[0, \frac{\pi}{2}\right)$, which is part of the principal value range for $\sec^{-1}$.

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

The principal value of $\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)$ is $\mathbf{\frac{\pi}{6}}$.

To Find:

The principal value of $\cot^{-1}(\sqrt{3})$.

Solution:

Let $y = \cot^{-1}(\sqrt{3})$.

By the definition of the inverse cotangent function, this means:

$\cot(y) = \sqrt{3}$

The principal value branch of $\cot^{-1}(x)$ has the range $(0, \pi)$.

We need to find the angle $y$ within this interval such that $\cot(y) = \sqrt{3}$.

We can think in terms of tangent: $\cot(y) = \sqrt{3}$ implies $\tan(y) = \frac{1}{\sqrt{3}}$.

We know from trigonometry that $\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$, which means $\cot\left(\frac{\pi}{6}\right) = \sqrt{3}$.

The angle $y = \frac{\pi}{6}$ lies within the principal value range $(0, \pi)$.

Therefore, the principal value of $\cot^{-1}(\sqrt{3})$ is $\frac{\pi}{6}$.

The principal value of $\cot^{-1}(\sqrt{3})$ is $\mathbf{\frac{\pi}{6}}$.

To Find:

The principal value of $\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)$.

Solution:

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

By the definition of the inverse cosine function, this means:

$\cos(y) = -\frac{1}{\sqrt{2}}$

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

We need to find the angle $y$ within this interval such that $\cos(y) = -\frac{1}{\sqrt{2}}$.

First, let's find the reference angle $\alpha$ in the first quadrant $[0, \frac{\pi}{2}]$ such that $\cos(\alpha) = \frac{1}{\sqrt{2}}$. We know that $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$. So, the reference angle is $\alpha = \frac{\pi}{4}$.

Since $\cos(y)$ is negative ($-\frac{1}{\sqrt{2}}$) and the principal value range is $[0, \pi]$, the angle $y$ must lie in the second quadrant ($\frac{\pi}{2} < y \le \pi$), where cosine is negative.

The angle in the second quadrant with reference angle $\alpha = \frac{\pi}{4}$ is given by:

$y = \pi - \alpha$

$y = \pi - \frac{\pi}{4}$

$y = \frac{4\pi - \pi}{4} = \frac{3\pi}{4}$

The angle $y = \frac{3\pi}{4}$ lies within the principal value range $[0, \pi]$.

Therefore, the principal value of $\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)$ is $\frac{3\pi}{4}$.

The principal value of $\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)$ is $\mathbf{\frac{3\pi}{4}}$.

To Find:

The principal value of $\text{cosec}^{-1}(-\sqrt{2})$.

Solution:

Let $y = \text{cosec}^{-1}(-\sqrt{2})$.

By the definition of the inverse cosecant function, this means:

$\text{cosec}(y) = -\sqrt{2}$

Since $\text{cosec}(y) = \frac{1}{\sin(y)}$, we have:

$\frac{1}{\sin(y)} = -\sqrt{2}$

$\implies \sin(y) = -\frac{1}{\sqrt{2}}$

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

We need to find the angle $y$ within this range such that $\sin(y) = -\frac{1}{\sqrt{2}}$.

First, consider the positive value. We know that $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.

Since $\sin(y)$ is negative and $y$ must be in the principal value range $\left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]$, the angle $y$ must lie in the interval $\left[-\frac{\pi}{2}, 0\right)$.

We use the property $\sin(-x) = -\sin(x)$.

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

The angle $y = -\frac{\pi}{4}$ lies within the principal value range $\left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]$. Specifically, it is in $\left[-\frac{\pi}{2}, 0\right)$.

Therefore, the principal value of $\text{cosec}^{-1}(-\sqrt{2})$ is $-\frac{\pi}{4}$.

The principal value of $\text{cosec}^{-1}(-\sqrt{2})$ is $\mathbf{-\frac{\pi}{4}}$.

To Find:

The value of the expression $\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)$.

Solution:

We need to find the principal values of each term in the expression and then add them.

Step 1: Find the principal value of $\tan^{-1}(1)$.

Let $y_1 = \tan^{-1}(1)$. Then $\tan(y_1) = 1$.

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

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

Since $\frac{\pi}{4}$ lies in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,

$\tan^{-1}(1) = \frac{\pi}{4}$.

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

Let $y_2 = \cos^{-1}\left(-\frac{1}{2}\right)$. Then $\cos(y_2) = -\frac{1}{2}$.

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

We know that $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$. Since $\cos(y_2)$ is negative, $y_2$ must lie in the second quadrant.

The angle in the second quadrant corresponding to the reference angle $\frac{\pi}{3}$ is $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

Since $\frac{2\pi}{3}$ lies in the interval $[0, \pi]$,

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

Step 3: Find the principal value of $\sin^{-1}\left(-\frac{1}{2}\right)$.

Let $y_3 = \sin^{-1}\left(-\frac{1}{2}\right)$. Then $\sin(y_3) = -\frac{1}{2}$.

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

We know that $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$. Since $\sin(y_3)$ is negative, $y_3$ must lie in the interval $\left[-\frac{\pi}{2}, 0\right]$.

Using $\sin(-x) = -\sin(x)$, we have $\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$.

Since $-\frac{\pi}{6}$ lies in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,

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

Step 4: Add the principal values.

The required value is $\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)$

$= \frac{\pi}{4} + \frac{2\pi}{3} + \left(-\frac{\pi}{6}\right)$

$= \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6}$

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

$= \frac{\pi \times 3}{4 \times 3} + \frac{2\pi \times 4}{3 \times 4} - \frac{\pi \times 2}{6 \times 2}$

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

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

$= \frac{11\pi - 2\pi}{12}$

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

Simplifying the fraction:

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

Therefore, the value of $\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)$ is $\mathbf{\frac{3\pi}{4}}$.

To Find:

The value of the expression $\cos^{-1}\left(\frac{1}{2}\right) + 2 \sin^{-1}\left(\frac{1}{2}\right)$.

Solution:

We need to find the principal values of each term in the expression and then combine them.

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

Let $y_1 = \cos^{-1}\left(\frac{1}{2}\right)$. Then $\cos(y_1) = \frac{1}{2}$.

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

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

Since $\frac{\pi}{3}$ lies in the interval $[0, \pi]$,

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

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

Let $y_2 = \sin^{-1}\left(\frac{1}{2}\right)$. Then $\sin(y_2) = \frac{1}{2}$.

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

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

Since $\frac{\pi}{6}$ lies in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,

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

Step 3: Calculate the value of the expression.

The required value is $\cos^{-1}\left(\frac{1}{2}\right) + 2 \sin^{-1}\left(\frac{1}{2}\right)$

$= \frac{\pi}{3} + 2 \times \left(\frac{\pi}{6}\right)$

$= \frac{\pi}{3} + \frac{2\pi}{6}$

Simplify the second term:

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

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

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

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

Given:

The relation $\sin^{-1}(x) = y$.

To Find:

The range of $y$ that represents the principal value branch of the inverse sine function.

Solution:

The function $y = \sin^{-1}(x)$ represents the inverse sine function. By definition, the range of the principal value branch of the inverse sine function is the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

This means that the value of $y$ must satisfy the inequality:

$-\frac{\pi}{2} \le y \le \frac{\pi}{2}$

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

(A) $0 \le y \le \pi$ : This is the principal value range for $\cos^{-1}(x)$.

(B) $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$ : This matches the principal value range for $\sin^{-1}(x)$.

(C) $0 < y < \pi$ : This is the principal value range for $\cot^{-1}(x)$.

(D) $-\frac{\pi}{2} < y < \frac{\pi}{2}$ : This is the principal value range for $\tan^{-1}(x)$.

Therefore, if $\sin^{-1}(x) = y$, the correct range for the principal value is $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$.

The correct option is (B).

To Find:

The value of the expression $\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2)$.

Solution:

We need to find the principal values of each term in the expression and then subtract the second from the first.

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

Let $y_1 = \tan^{-1}(\sqrt{3})$. Then $\tan(y_1) = \sqrt{3}$.

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

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

Since $\frac{\pi}{3}$ lies in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,

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

Step 2: Find the principal value of $\sec^{-1}(-2)$.

Let $y_2 = \sec^{-1}(-2)$. Then $\sec(y_2) = -2$.

This implies $\cos(y_2) = \frac{1}{\sec(y_2)} = -\frac{1}{2}$.

The range of the principal value branch of $\sec^{-1}(x)$ is $\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]$. Alternatively, we find the angle $y_2$ in $[0, \pi]$ such that $\cos(y_2) = -\frac{1}{2}$ and check if $y_2 \neq \frac{\pi}{2}$.

We know that $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$. Since $\cos(y_2)$ is negative, $y_2$ must lie in the second quadrant within the range $[0, \pi]$.

The angle in the second quadrant corresponding to the reference angle $\frac{\pi}{3}$ is $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

Since $\frac{2\pi}{3}$ lies in the interval $[0, \pi]$ and $\frac{2\pi}{3} \neq \frac{\pi}{2}$, it is in the principal value range for $\sec^{-1}(x)$.

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

Step 3: Calculate the value of the expression.

The required value is $\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2)$

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

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

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

Comparing this value with the given options:

(A) $\pi$

(B) $-\frac{\pi}{3}$

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

(D) $\frac{2\pi}{3}$

The calculated value matches option (B).

Therefore, $\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2)$ is equal to $\mathbf{-\frac{\pi}{3}}$.

The correct option is (B).

To Show:

(i) $\sin^{-1}\left( 2x\sqrt{1-x^{2}} \right) = 2 \sin^{-1} x$, for $-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}$

(ii) $\sin^{-1}\left( 2x\sqrt{1-x^{2}} \right) = 2 \cos^{-1} x$, for $\frac{1}{\sqrt{2}} \le x \le 1$

Proof (i):

Let $x = \sin\theta$.

Given the range $-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}$, we have:

$-\frac{1}{\sqrt{2}} \le \sin\theta \le \frac{1}{\sqrt{2}}$

Since the principal value range of $\sin^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, this implies:

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

$-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}$

Now consider the expression inside the $\sin^{-1}$ function on the LHS:

$2x\sqrt{1-x^2} = 2\sin\theta\sqrt{1-\sin^2\theta}$

$= 2\sin\theta\sqrt{\cos^2\theta}$

$= 2\sin\theta|\cos\theta|$

Since $-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}$, $\theta$ is in the first or fourth quadrant, where $\cos\theta \ge 0$. Thus, $|\cos\theta| = \cos\theta$.

$= 2\sin\theta\cos\theta$

Using the double angle identity, $\sin(2\theta) = 2\sin\theta\cos\theta$.

$= \sin(2\theta)$

So, the LHS becomes $\sin^{-1}(\sin(2\theta))$.

We need to check the range of $2\theta$. Since $-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}$, multiplying by 2 gives:

$-\frac{\pi}{2} \le 2\theta \le \frac{\pi}{2}$

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

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

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

LHS $= 2\theta = 2\sin^{-1}x$.

This is equal to the RHS.

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

Proof (ii):

Let $x = \cos\theta$.

Given the range $\frac{1}{\sqrt{2}} \le x \le 1$, we have:

$\frac{1}{\sqrt{2}} \le \cos\theta \le 1$

Since the principal value range of $\cos^{-1}$ is $[0, \pi]$, this implies:

$\cos^{-1}(1) \le \theta \le \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)$ (Note: $\cos^{-1}$ is a decreasing function)

$0 \le \theta \le \frac{\pi}{4}$

Now consider the expression inside the $\sin^{-1}$ function on the LHS:

$2x\sqrt{1-x^2} = 2\cos\theta\sqrt{1-\cos^2\theta}$

$= 2\cos\theta\sqrt{\sin^2\theta}$

$= 2\cos\theta|\sin\theta|$

Since $0 \le \theta \le \frac{\pi}{4}$, $\theta$ is in the first quadrant, where $\sin\theta \ge 0$. Thus, $|\sin\theta| = \sin\theta$.

$= 2\sin\theta\cos\theta$

Using the double angle identity, $\sin(2\theta) = 2\sin\theta\cos\theta$.

$= \sin(2\theta)$

So, the LHS becomes $\sin^{-1}(\sin(2\theta))$.

We need to check the range of $2\theta$. Since $0 \le \theta \le \frac{\pi}{4}$, multiplying by 2 gives:

$0 \le 2\theta \le \frac{\pi}{2}$

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

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

Substitute back $\theta = \cos^{-1}x$:

LHS $= 2\theta = 2\cos^{-1}x$.

This is equal to the RHS.

Hence, $\sin^{-1}\left( 2x\sqrt{1-x^{2}} \right) = 2 \cos^{-1} x$ for $\frac{1}{\sqrt{2}} \le x \le 1$.

To Show:

$\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{2}{11}\right) = \tan^{-1}\left(\frac{3}{4}\right)$

Proof:

We use the identity for the sum of two inverse tangent functions:

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

Let $x = \frac{1}{2}$ and $y = \frac{2}{11}$.

First, we check the condition $xy < 1$:

$xy = \left(\frac{1}{2}\right) \times \left(\frac{2}{11}\right) = \frac{2}{22} = \frac{1}{11}$.

Since $\frac{1}{11} < 1$, the condition is satisfied, and we can apply the identity.

Now, consider the Left Hand Side (LHS):

LHS $= \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{2}{11}\right)$

$= \tan^{-1}\left(\frac{\frac{1}{2} + \frac{2}{11}}{1 - \left(\frac{1}{2}\right)\left(\frac{2}{11}\right)}\right)$

$= \tan^{-1}\left(\frac{\frac{11 + 4}{22}}{1 - \frac{2}{22}}\right)$

$= \tan^{-1}\left(\frac{\frac{15}{22}}{1 - \frac{1}{11}}\right)$

$= \tan^{-1}\left(\frac{\frac{15}{22}}{\frac{11 - 1}{11}}\right)$

$= \tan^{-1}\left(\frac{\frac{15}{22}}{\frac{10}{11}}\right)$

$= \tan^{-1}\left(\frac{15}{22} \times \frac{11}{10}\right)$

$= \tan^{-1}\left(\frac{15 \times \cancel{11}^1}{\cancel{22}_2 \times 10}\right)$

$= \tan^{-1}\left(\frac{15}{20}\right)$

Simplify the fraction by dividing the numerator and denominator by 5:

$= \tan^{-1}\left(\frac{\cancel{15}^3}{\cancel{20}_4}\right)$

$= \tan^{-1}\left(\frac{3}{4}\right)$

This is equal to the Right Hand Side (RHS).

LHS = $\tan^{-1}\left(\frac{3}{4}\right)$ = RHS.

Hence, it is shown that $\mathbf{\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{2}{11}\right) = \tan^{-1}\left(\frac{3}{4}\right)}$.

Given:

The expression $\tan^{-1}\left( \frac{\cos x}{1-\sin x} \right)$ for $-\frac{3\pi}{2} < x < \frac{\pi}{2}$.

To Find:

The simplest form of the given expression.

Solution:

We want to simplify the argument of the $\tan^{-1}$ function, $\frac{\cos x}{1-\sin x}$.

We use the half-angle identities:

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

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

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

Substitute these into the fraction:

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

Factor the numerator as a difference of squares and the denominator as a perfect square:

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

We need to ensure the term we cancel is not zero. $\cos\left(\frac{x}{2}\right) - \sin\left(\frac{x}{2}\right) = 0$ implies $\tan\left(\frac{x}{2}\right) = 1$, so $\frac{x}{2} = n\pi + \frac{\pi}{4}$, which gives $x = 2n\pi + \frac{\pi}{2}$. The value $x = \frac{\pi}{2}$ is not included in the interval $-\frac{3\pi}{2} < x < \frac{\pi}{2}$. The value $x = -2\pi + \frac{\pi}{2} = -\frac{3\pi}{2}$ is also not included. Thus, we can cancel the term.

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

Divide the numerator and the denominator by $\cos\left(\frac{x}{2}\right)$ (assuming $\cos\left(\frac{x}{2}\right) \neq 0$, which means $x \neq \pi + 2n\pi$; we will check these cases later if needed):

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

Recall the tangent addition formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$. Using $A = \frac{\pi}{4}$ and $B = \frac{x}{2}$, we have $\tan\left(\frac{\pi}{4}\right) = 1$.

$\frac{1 + \tan\left(\frac{x}{2}\right)}{1 - \tan\left(\frac{x}{2}\right)} = \frac{\tan\left(\frac{\pi}{4}\right) + \tan\left(\frac{x}{2}\right)}{1 - \tan\left(\frac{\pi}{4}\right)\tan\left(\frac{x}{2}\right)} = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right)$

So the original expression becomes:

$\tan^{-1}\left(\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\right)$

We can simplify $\tan^{-1}(\tan \theta) = \theta$ if $\theta$ lies in the principal value range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Let's check the range of $\theta = \frac{\pi}{4} + \frac{x}{2}$.

Given $-\frac{3\pi}{2} < x < \frac{\pi}{2}$.

Divide by 2: $-\frac{3\pi}{4} < \frac{x}{2} < \frac{\pi}{4}$.

Add $\frac{\pi}{4}$: $-\frac{3\pi}{4} + \frac{\pi}{4} < \frac{x}{2} + \frac{\pi}{4} < \frac{\pi}{4} + \frac{\pi}{4}$.

$-\frac{2\pi}{4} < \frac{\pi}{4} + \frac{x}{2} < \frac{2\pi}{4}$.

$-\frac{\pi}{2} < \frac{\pi}{4} + \frac{x}{2} < \frac{\pi}{2}$.

Since the argument $\frac{\pi}{4} + \frac{x}{2}$ lies in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, the simplification $\tan^{-1}(\tan \theta) = \theta$ is valid.

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

(Checking the case where $\cos(x/2) = 0$: This occurs at $x = \pi, -\pi$. $x=-\pi$ is in the domain. For $x = -\pi$, the original expression is $\tan^{-1}(\frac{\cos(-\pi)}{1-\sin(-\pi)}) = \tan^{-1}(\frac{-1}{1-0}) = \tan^{-1}(-1) = -\frac{\pi}{4}$. The simplified form gives $\frac{\pi}{4} + \frac{-\pi}{2} = -\frac{\pi}{4}$. They match.)

The simplest form of the expression is $\mathbf{\frac{\pi}{4} + \frac{x}{2}}$.

Given:

The expression $\cot^{-1}\left( \frac{1}{\sqrt{x^{2}-1}} \right)$, where $x > 1$.

To Find:

The simplest form of the given expression.

Solution:

The presence of the term $\sqrt{x^2 - 1}$ suggests a trigonometric substitution.

Let $x = \sec\theta$.

Since $x > 1$, we have $\sec\theta > 1$. This implies that $\theta$ can be chosen in the interval $\left(0, \frac{\pi}{2}\right)$. In this interval, $\tan\theta$ is positive.

Now, substitute $x = \sec\theta$ into the expression $\sqrt{x^2 - 1}$:

$\sqrt{x^2 - 1} = \sqrt{\sec^2\theta - 1}$

Using the identity $\sec^2\theta - 1 = \tan^2\theta$:

$\sqrt{x^2 - 1} = \sqrt{\tan^2\theta}$

Since $\theta \in \left(0, \frac{\pi}{2}\right)$, $\tan\theta > 0$. Therefore, $\sqrt{\tan^2\theta} = \tan\theta$.

Substitute this back into the original expression:

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

Using the identity $\frac{1}{\tan\theta} = \cot\theta$:

$= \cot^{-1}(\cot\theta)$

The principal value range for $\cot^{-1}(y)$ is $(0, \pi)$.

We chose $\theta \in \left(0, \frac{\pi}{2}\right)$. Since this interval is within $(0, \pi)$, we can simplify $\cot^{-1}(\cot\theta)$ to $\theta$.

So, $\cot^{-1}(\cot\theta) = \theta$.

Now, we need to express $\theta$ back in terms of $x$.

Since we let $x = \sec\theta$, it follows that $\theta = \sec^{-1}x$.

Therefore, the simplest form of the given expression is $\sec^{-1}x$.

The simplest form of $\cot^{-1}\left( \frac{1}{\sqrt{x^{2}-1}} \right)$ for $x > 1$ is $\mathbf{\sec^{-1}x}$.

To Prove:

$\tan^{-1} x + \tan^{-1} \left(\frac{2x}{1-x^{2}}\right) = \tan^{-1} \left( \frac{3x-x^{3}}{1-3x^{2}} \right)$, for $|x| < \frac{1}{\sqrt{3}}$.

Proof:

Let $x = \tan\theta$.

The condition $|x| < \frac{1}{\sqrt{3}}$ translates to $|\tan\theta| < \frac{1}{\sqrt{3}}$.

Since the principal value range of $\tan^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, we have:

$\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) < \theta < \tan^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$-\frac{\pi}{6} < \theta < \frac{\pi}{6}$

Now consider the Left Hand Side (LHS) of the equation:

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

Substitute $x = \tan\theta$:

LHS $= \tan^{-1}(\tan\theta) + \tan^{-1}\left(\frac{2\tan\theta}{1-\tan^2\theta}\right)$

Using the double angle identity $\tan(2\alpha) = \frac{2\tan\alpha}{1-\tan^2\alpha}$:

LHS $= \tan^{-1}(\tan\theta) + \tan^{-1}(\tan(2\theta))$

We need to evaluate these terms using the property $\tan^{-1}(\tan\alpha) = \alpha$ if $\alpha$ is in the principal range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

For the first term, $\tan^{-1}(\tan\theta)$: Since $-\frac{\pi}{6} < \theta < \frac{\pi}{6}$, $\theta$ is within the principal range. So, $\tan^{-1}(\tan\theta) = \theta$.

For the second term, $\tan^{-1}(\tan(2\theta))$: We need the range of $2\theta$. Since $-\frac{\pi}{6} < \theta < \frac{\pi}{6}$, multiplying by 2 gives $-\frac{\pi}{3} < 2\theta < \frac{\pi}{3}$. This range is also within the principal range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. So, $\tan^{-1}(\tan(2\theta)) = 2\theta$.

Therefore, LHS $= \theta + 2\theta = 3\theta$.

Now consider the Right Hand Side (RHS) of the equation:

RHS $= \tan^{-1}\left( \frac{3x-x^{3}}{1-3x^{2}} \right)$

Substitute $x = \tan\theta$:

RHS $= \tan^{-1}\left( \frac{3\tan\theta - \tan^3\theta}{1-3\tan^2\theta} \right)$

Using the triple angle identity $\tan(3\alpha) = \frac{3\tan\alpha - \tan^3\alpha}{1-3\tan^2\alpha}$:

RHS $= \tan^{-1}(\tan(3\theta))$

We need to check the range of $3\theta$. Since $-\frac{\pi}{6} < \theta < \frac{\pi}{6}$, multiplying by 3 gives $-\frac{\pi}{2} < 3\theta < \frac{\pi}{2}$. This range is exactly the principal range for $\tan^{-1}$.

Therefore, $\tan^{-1}(\tan(3\theta)) = 3\theta$.

We have shown that LHS $= 3\theta$ and RHS $= 3\theta$.

Thus, LHS = RHS.

Hence, it is proved that $\mathbf{\tan^{-1} x + \tan^{-1} \left(\frac{2x}{1-x^{2}}\right) = \tan^{-1} \left( \frac{3x-x^{3}}{1-3x^{2}} \right)}$ for $|x| < \frac{1}{\sqrt{3}}$.

Given:

The expression $\cos(\sec^{-1}x + \text{cosec}^{-1}x)$, where $|x| \ge 1$.

To Find:

The value of the given expression.

Solution:

We know the identity relating the inverse secant and inverse cosecant functions:

$\sec^{-1}x + \text{cosec}^{-1}x = \frac{\pi}{2}$

This identity holds for all values of $x$ for which both $\sec^{-1}x$ and $\text{cosec}^{-1}x$ are defined. The domain for both functions is $(-\infty, -1] \cup [1, \infty)$, which is equivalent to $|x| \ge 1$.

Since the problem states that $|x| \ge 1$, the condition for the identity is met.

Now, substitute the identity into the given expression:

$\cos(\sec^{-1}x + \text{cosec}^{-1}x) = \cos\left(\frac{\pi}{2}\right)$

We know the value of $\cos\left(\frac{\pi}{2}\right)$ is 0.

Therefore, $\cos\left(\frac{\pi}{2}\right) = 0$.

The value of $\cos(\sec^{-1}x + \text{cosec}^{-1}x)$ for $|x| \ge 1$ is 0.

To Prove:

$3 \sin^{-1} x = \sin^{-1} (3x - 4x^3)$, for $x \in \left[ -\frac{1}{2} , \frac{1}{2}\right]$.

Proof:

Let $x = \sin\theta$.

Since $x \in \left[ -\frac{1}{2} , \frac{1}{2}\right]$, we have:

$-\frac{1}{2} \le \sin\theta \le \frac{1}{2}$

Considering the principal value range of $\sin^{-1}(x)$ which is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, taking $\sin^{-1}$ of the inequality gives:

$\sin^{-1}\left(-\frac{1}{2}\right) \le \theta \le \sin^{-1}\left(\frac{1}{2}\right)$

$-\frac{\pi}{6} \le \theta \le \frac{\pi}{6}$

Now consider the Right Hand Side (RHS) of the equation:

RHS $= \sin^{-1}(3x - 4x^3)$

Substitute $x = \sin\theta$:

RHS $= \sin^{-1}(3\sin\theta - 4\sin^3\theta)$

Using the triple angle identity for sine, $\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$:

RHS $= \sin^{-1}(\sin(3\theta))$

Now, we need to check the range of $3\theta$.

Since $-\frac{\pi}{6} \le \theta \le \frac{\pi}{6}$, multiplying by 3 gives:

$3 \times \left(-\frac{\pi}{6}\right) \le 3\theta \le 3 \times \left(\frac{\pi}{6}\right)$

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

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

Therefore, $\sin^{-1}(\sin(3\theta)) = 3\theta$.

RHS $= 3\theta$

Now substitute back $\theta = \sin^{-1}x$:

RHS $= 3\sin^{-1}x$.

The Left Hand Side (LHS) is $3\sin^{-1}x$.

Since RHS = LHS, the identity is proved.

Hence, it is proved that $\mathbf{3 \sin^{-1} x = \sin^{-1} (3x - 4x^3)}$ for $x \in \left[ -\frac{1}{2} , \frac{1}{2}\right]$.

To Prove:

$3 \cos^{-1} x = \cos^{-1} (4x^3 - 3x)$, for $x \in \left[ \frac{1}{2} , 1 \right]$.

Proof:

Let $x = \cos\theta$.

Since $x \in \left[ \frac{1}{2} , 1 \right]$, we have:

$\frac{1}{2} \le \cos\theta \le 1$

Considering the principal value range of $\cos^{-1}(x)$ which is $[0, \pi]$, taking $\cos^{-1}$ of the inequality (and reversing the inequality signs because $\cos^{-1}$ is a decreasing function):

$\cos^{-1}(1) \le \theta \le \cos^{-1}\left(\frac{1}{2}\right)$

$0 \le \theta \le \frac{\pi}{3}$

Now consider the Right Hand Side (RHS) of the equation:

RHS $= \cos^{-1}(4x^3 - 3x)$

Substitute $x = \cos\theta$:

RHS $= \cos^{-1}(4\cos^3\theta - 3\cos\theta)$

Using the triple angle identity for cosine, $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$:

RHS $= \cos^{-1}(\cos(3\theta))$

Now, we need to check the range of $3\theta$.

Since $0 \le \theta \le \frac{\pi}{3}$, multiplying by 3 gives:

$3 \times 0 \le 3\theta \le 3 \times \frac{\pi}{3}$

$0 \le 3\theta \le \pi$

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

Therefore, $\cos^{-1}(\cos(3\theta)) = 3\theta$.

RHS $= 3\theta$

Now substitute back $\theta = \cos^{-1}x$:

RHS $= 3\cos^{-1}x$.

The Left Hand Side (LHS) is $3\cos^{-1}x$.

Since RHS = LHS, the identity is proved.

Hence, it is proved that $\mathbf{3 \cos^{-1} x = \cos^{-1} (4x^3 - 3x)}$ for $x \in \left[ \frac{1}{2} , 1 \right]$.

To Prove:

$\tan^{-1}\left(\frac{2}{11}\right) + \tan^{-1}\left(\frac{7}{24}\right) = \tan^{-1}\left(\frac{1}{2}\right)$

Proof:

We start with the Left Hand Side (LHS) and use the identity for the sum of two inverse tangent functions:

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

Let $x = \frac{2}{11}$ and $y = \frac{7}{24}$.

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

$xy = \left(\frac{2}{11}\right) \times \left(\frac{7}{24}\right) = \frac{2 \times 7}{11 \times 24} = \frac{14}{264}$.

Since $14 < 264$, we have $xy < 1$. The condition is satisfied, so we can apply the identity.

LHS $= \tan^{-1}\left(\frac{2}{11}\right) + \tan^{-1}\left(\frac{7}{24}\right)$

$= \tan^{-1}\left(\frac{\frac{2}{11} + \frac{7}{24}}{1 - \left(\frac{2}{11}\right)\left(\frac{7}{24}\right)}\right)$

$= \tan^{-1}\left(\frac{\frac{2 \times 24 + 7 \times 11}{11 \times 24}}{1 - \frac{14}{264}}\right)$

$= \tan^{-1}\left(\frac{\frac{48 + 77}{264}}{\frac{264 - 14}{264}}\right)$

$= \tan^{-1}\left(\frac{\frac{125}{264}}{\frac{250}{264}}\right)$

We can cancel the denominator 264 from the numerator and denominator of the main fraction:

$= \tan^{-1}\left(\frac{125}{250}\right)$

Simplify the fraction $\frac{125}{250}$:

$= \tan^{-1}\left(\frac{\cancel{125}^1}{\cancel{250}_2}\right)$

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

This is equal to the Right Hand Side (RHS).

LHS = $\tan^{-1}\left(\frac{1}{2}\right)$ = RHS.

Hence, it is proved that $\mathbf{\tan^{-1}\left(\frac{2}{11}\right) + \tan^{-1}\left(\frac{7}{24}\right) = \tan^{-1}\left(\frac{1}{2}\right)}$.

To Prove:

$2\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{7}\right) = \tan^{-1}\left(\frac{31}{17}\right)$

Proof:

We start with the Left Hand Side (LHS): $2\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{7}\right)$.

First, we simplify the term $2\tan^{-1}\left(\frac{1}{2}\right)$ using the identity:

$2\tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$, provided $|x| < 1$.

Here, $x = \frac{1}{2}$. Since $|x| = \frac{1}{2} < 1$, the identity is applicable.

$2\tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{2 \times \frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2}\right)$

$= \tan^{-1}\left(\frac{1}{1 - \frac{1}{4}}\right)$

$= \tan^{-1}\left(\frac{1}{\frac{3}{4}}\right)$

$= \tan^{-1}\left(\frac{4}{3}\right)$

Now substitute this back into the LHS:

LHS $= \tan^{-1}\left(\frac{4}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right)$

Next, we use the identity for the sum of two inverse tangent functions:

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

Let $x = \frac{4}{3}$ and $y = \frac{1}{7}$.

Check the condition $xy < 1$.

$xy = \left(\frac{4}{3}\right) \times \left(\frac{1}{7}\right) = \frac{4}{21}$.

Since $4 < 21$, we have $xy < 1$. The condition is satisfied, so we can apply the identity.

LHS $= \tan^{-1}\left(\frac{\frac{4}{3} + \frac{1}{7}}{1 - \left(\frac{4}{3}\right)\left(\frac{1}{7}\right)}\right)$

$= \tan^{-1}\left(\frac{\frac{4 \times 7 + 1 \times 3}{3 \times 7}}{1 - \frac{4}{21}}\right)$

$= \tan^{-1}\left(\frac{\frac{28 + 3}{21}}{\frac{21 - 4}{21}}\right)$

$= \tan^{-1}\left(\frac{\frac{31}{21}}{\frac{17}{21}}\right)$

Cancel the denominator 21:

$= \tan^{-1}\left(\frac{31}{17}\right)$

This is equal to the Right Hand Side (RHS).

LHS = $\tan^{-1}\left(\frac{31}{17}\right)$ = RHS.

Hence, it is proved that $\mathbf{2\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{7}\right) = \tan^{-1}\left(\frac{31}{17}\right)}$.

Given:

The function $\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$, where $x \neq 0$.

To Find:

The simplest form of the given function.

Solution:

The presence of the term $\sqrt{1+x^2}$ suggests the trigonometric substitution $x = \tan\theta$.

Let $x = \tan\theta$. Since the range of $\tan^{-1}x$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, we can assume $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

The condition $x \neq 0$ implies $\tan\theta \neq 0$, so $\theta \neq 0$.

Now substitute $x = \tan\theta$ into the expression inside the $\tan^{-1}$ function:

$\frac{\sqrt{1+x^2}-1}{x} = \frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta}$

Using the identity $1 + \tan^2\theta = \sec^2\theta$:

$= \frac{\sqrt{\sec^2\theta}-1}{\tan\theta}$

$= \frac{|\sec\theta|-1}{\tan\theta}$

For $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, $\sec\theta = \frac{1}{\cos\theta}$ is always positive. Therefore, $|\sec\theta| = \sec\theta$.

$= \frac{\sec\theta-1}{\tan\theta}$

Convert $\sec\theta$ and $\tan\theta$ to $\sin\theta$ and $\cos\theta$:

$= \frac{\frac{1}{\cos\theta}-1}{\frac{\sin\theta}{\cos\theta}}$

$= \frac{\frac{1-\cos\theta}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}}$

Multiply the numerator and denominator by $\cos\theta$ (since $\theta \neq \pm \frac{\pi}{2}$, $\cos\theta \neq 0$):

$= \frac{1-\cos\theta}{\sin\theta}$

Now, use the half-angle identities:

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

$\sin\theta = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)$

Substitute these into the expression:

$= \frac{2\sin^2\left(\frac{\theta}{2}\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)}$

Since $\theta \neq 0$, $\frac{\theta}{2} \neq 0$, which means $\sin\left(\frac{\theta}{2}\right) \neq 0$. We can cancel the term $2\sin\left(\frac{\theta}{2}\right)$:

$= \frac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)}$

$= \tan\left(\frac{\theta}{2}\right)$

So the original expression becomes:

$\tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right)$

We need to check the range of $\frac{\theta}{2}$. Since $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and $\theta \neq 0$, we have $\frac{\theta}{2} \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$ and $\frac{\theta}{2} \neq 0$.

This range $\left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \setminus \{0\}$ is within the principal value range of $\tan^{-1}(y)$, which is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Therefore, we can simplify $\tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right) = \frac{\theta}{2}$.

Finally, substitute back $\theta = \tan^{-1}x$.

The simplified expression is $\frac{1}{2}\theta = \frac{1}{2}\tan^{-1}x$.

The simplest form of $\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ for $x \neq 0$ is $\mathbf{\frac{1}{2}\tan^{-1}x}$.

Given:

The function $\tan^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)$, where $|x| > 1$.

To Find:

The simplest form of the given function.

Solution:

The presence of the term $\sqrt{x^2-1}$ suggests a trigonometric substitution.

Let $x = \text{cosec}\phi$.

The condition $|x| > 1$ means $x > 1$ or $x < -1$.

The principal value range for $\phi = \text{cosec}^{-1}x$ is $\left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]$.

- If $x > 1$, then $\phi \in \left(0, \frac{\pi}{2}\right)$.

- If $x < -1$, then $\phi \in \left(-\frac{\pi}{2}, 0\right)$.

Now substitute $x = \text{cosec}\phi$ into the term $\sqrt{x^2-1}$:

$\sqrt{x^2-1} = \sqrt{\text{cosec}^2\phi - 1}$

Using the identity $\text{cosec}^2\phi - 1 = \cot^2\phi$:

$\sqrt{x^2-1} = \sqrt{\cot^2\phi} = |\cot\phi|$.

The expression becomes:

$E = \tan^{-1}\left(\frac{1}{|\cot\phi|}\right)$

Since $\frac{1}{\cot\phi} = \tan\phi$, we have $\frac{1}{|\cot\phi|} = |\tan\phi|$.

$E = \tan^{-1}(|\tan\phi|)$.

Now we consider the two cases for $x$:

Case 1: $x > 1$

In this case, $\phi \in \left(0, \frac{\pi}{2}\right)$. For angles in this interval, $\tan\phi > 0$.

Therefore, $|\tan\phi| = \tan\phi$.

The expression becomes $E = \tan^{-1}(\tan\phi)$.

Since $\phi \in \left(0, \frac{\pi}{2}\right)$, which is within the principal value range of $\tan^{-1}(y)$, $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, we have:

$E = \phi$.

Substituting back $\phi = \text{cosec}^{-1}x$, we get:

$E = \text{cosec}^{-1}x$ for $x > 1$.

Case 2: $x < -1$

In this case, $\phi \in \left(-\frac{\pi}{2}, 0\right)$. For angles in this interval, $\tan\phi < 0$.

Therefore, $|\tan\phi| = -\tan\phi$.

The expression becomes $E = \tan^{-1}(-\tan\phi)$.

Using the property $\tan^{-1}(-y) = -\tan^{-1}(y)$:

$E = -\tan^{-1}(\tan\phi)$.

Since $\phi \in \left(-\frac{\pi}{2}, 0\right)$, which is within the principal value range of $\tan^{-1}(y)$, $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, we have $\tan^{-1}(\tan\phi) = \phi$.

$E = -\phi$.

Substituting back $\phi = \text{cosec}^{-1}x$, we get:

$E = -\text{cosec}^{-1}x$ for $x < -1$.

Conclusion:

The simplest form of the function is:

$\tan^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right) = \begin{cases} \text{cosec}^{-1}x & , & \text{if } x > 1 \\ -\text{cosec}^{-1}x & , & \text{if } x < -1 \end{cases}$

Alternatively, using the identity $\text{cosec}^{-1}x = \frac{\pi}{2} - \sec^{-1}x$ (for $|x|\ge 1$), we can write this as:

$\begin{cases} \frac{\pi}{2} - \sec^{-1}x & , & \text{if } x > 1 \\ -(\frac{\pi}{2} - \sec^{-1}x) = \sec^{-1}x - \frac{\pi}{2} & , & \text{if } x < -1 \end{cases}$

Given:

The function $\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)$, where $0 < x < \pi$.

To Find:

The simplest form of the given function.

Solution:

First, we simplify the expression inside the square root using half-angle identities:

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

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

Substitute these into the fraction:

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

Now, substitute this back into the argument of the $\tan^{-1}$ function:

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

$= \left|\tan\left(\frac{x}{2}\right)\right|$

We need to determine the sign of $\tan\left(\frac{x}{2}\right)$ based on the given range for $x$: $0 < x < \pi$.

Dividing the inequality by 2, we get:

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

This means the angle $\frac{x}{2}$ lies in the first quadrant.

In the first quadrant, the tangent function is positive. Therefore, $\tan\left(\frac{x}{2}\right) > 0$.

So, $\left|\tan\left(\frac{x}{2}\right)\right| = \tan\left(\frac{x}{2}\right)$.

The original expression simplifies to:

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

We can use the property $\tan^{-1}(\tan \theta) = \theta$ if $\theta$ lies within the principal value range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

We found that the range for $\theta = \frac{x}{2}$ is $\left(0, \frac{\pi}{2}\right)$.

Since $\left(0, \frac{\pi}{2}\right)$ is contained within $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, the property applies.

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

The simplest form of $\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)$ for $0 < x < \pi$ is $\mathbf{\frac{x}{2}}$.

Given:

The function $\tan^{-1}\left( \frac{\cos x - \sin x}{\cos x + \sin x} \right)$, where $-\frac{\pi}{4} < x < \frac{3\pi}{4}$.

To Find:

The simplest form of the given function.

Solution:

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

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

Divide both the numerator and the denominator by $\cos x$. We need to consider if $\cos x$ can be zero in the given interval. $\cos x = 0$ when $x = \frac{\pi}{2} + n\pi$. For the interval $-\frac{\pi}{4} < x < \frac{3\pi}{4}$, the only value where $\cos x = 0$ is $x = \frac{\pi}{2}$. Let's proceed with the division, assuming $\cos x \neq 0$ for now, and check the case $x=\frac{\pi}{2}$ later.

$E = \frac{\frac{\cos x}{\cos x} - \frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x} + \frac{\sin x}{\cos x}} = \frac{1 - \tan x}{1 + \tan x}$

We can rewrite this expression using the tangent subtraction formula, $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.

Let $A = \frac{\pi}{4}$, so $\tan A = 1$.

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

So the original expression becomes:

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

We can use the property $\tan^{-1}(\tan \theta) = \theta$ if $\theta$ lies within the principal value range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

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

We are given $-\frac{\pi}{4} < x < \frac{3\pi}{4}$.

Multiply by -1 (reversing the inequalities):

$-\frac{3\pi}{4} < -x < \frac{\pi}{4}$

Add $\frac{\pi}{4}$ to all parts:

$\frac{\pi}{4} - \frac{3\pi}{4} < \frac{\pi}{4} - x < \frac{\pi}{4} + \frac{\pi}{4}$

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

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

Since the angle $\theta = \frac{\pi}{4} - x$ lies within the principal value range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, the property applies.

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

Now, let's check the case $x = \frac{\pi}{2}$, where $\cos x = 0$.

Original expression: $\tan^{-1}\left( \frac{\cos(\pi/2) - \sin(\pi/2)}{\cos(\pi/2) + \sin(\pi/2)} \right) = \tan^{-1}\left( \frac{0 - 1}{0 + 1} \right) = \tan^{-1}(-1) = -\frac{\pi}{4}$.

Simplified form: $\frac{\pi}{4} - x = \frac{\pi}{4} - \frac{\pi}{2} = -\frac{\pi}{4}$.

The simplified form holds even for $x = \frac{\pi}{2}$.

The simplest form of $\tan^{-1}\left( \frac{\cos x - \sin x}{\cos x + \sin x} \right)$ for $-\frac{\pi}{4} < x < \frac{3\pi}{4}$ is $\mathbf{\frac{\pi}{4} - x}$.

Given:

The function $\tan^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)$, where $|x| < a$. Assume $a > 0$ for the expression $\sqrt{a^2-x^2}$ to be defined for a range of $x$.

To Find:

The simplest form of the given function.

Solution:

The term $\sqrt{a^2-x^2}$ suggests the trigonometric substitution $x = a\sin\theta$.

The condition $|x| < a$ translates to $-a < x < a$.

Substituting $x = a\sin\theta$ gives $-a < a\sin\theta < a$.

Since we assume $a > 0$, we can divide by $a$: $-1 < \sin\theta < 1$.

This means we can choose $\theta$ within the principal value range of the inverse sine function, excluding the endpoints: $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. From $x = a\sin\theta$, we get $\sin\theta = \frac{x}{a}$, so $\theta = \sin^{-1}\left(\frac{x}{a}\right)$.

Now, substitute $x = a\sin\theta$ into the expression $\frac{x}{\sqrt{a^2-x^2}}$:

Numerator: $x = a\sin\theta$.

Denominator: $\sqrt{a^2-x^2} = \sqrt{a^2 - (a\sin\theta)^2} = \sqrt{a^2 - a^2\sin^2\theta}$

$= \sqrt{a^2(1-\sin^2\theta)} = \sqrt{a^2\cos^2\theta}$

$= \sqrt{(a\cos\theta)^2} = |a\cos\theta|$

Since $a > 0$, $|a\cos\theta| = a|\cos\theta|$.

For the range $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, the cosine function is positive, $\cos\theta > 0$.

Therefore, $|\cos\theta| = \cos\theta$, and the denominator is $a\cos\theta$.

The fraction inside the $\tan^{-1}$ function becomes:

$\frac{x}{\sqrt{a^2-x^2}} = \frac{a\sin\theta}{a\cos\theta}$

Since $|x| < a$ and $a>0$, $a \neq 0$. We can cancel $a$.

$\frac{\sin\theta}{\cos\theta} = \tan\theta$.

So the original function is:

$\tan^{-1}(\tan\theta)$

We can use the property $\tan^{-1}(\tan y) = y$ if $y$ lies in the principal value range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

In our case, $y = \theta$, and we found that $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

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

Finally, substitute back $\theta = \sin^{-1}\left(\frac{x}{a}\right)$.

The simplest form of $\tan^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)$ for $|x| < a$ is $\mathbf{\sin^{-1}\left(\frac{x}{a}\right)}$.

Given:

The function $\tan^{-1}\left( \frac{3a^{2}x-x^{3}}{a^{3}-3ax^{2}} \right)$, where $a > 0$ and $\frac{-a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$.

To Find:

The simplest form of the given function.

Solution:

The expression inside the $\tan^{-1}$ resembles the tangent triple angle formula $\tan(3\theta) = \frac{3\tan\theta - \tan^3\theta}{1-3\tan^2\theta}$.

Let's manipulate the expression $\frac{3a^{2}x-x^{3}}{a^{3}-3ax^{2}}$. Divide the numerator and denominator by $a^3$ (since $a>0$, $a^3 \neq 0$):

$\frac{\frac{3a^{2}x}{a^3}-\frac{x^{3}}{a^3}}{\frac{a^{3}}{a^3}-\frac{3ax^{2}}{a^3}} = \frac{3\left(\frac{x}{a}\right) - \left(\frac{x}{a}\right)^3}{1 - 3\left(\frac{x}{a}\right)^2}$

This suggests the substitution $x = a\tan\theta$, which implies $\frac{x}{a} = \tan\theta$.

Now, let's determine the range for $\theta$ using the given condition on $x$: $\frac{-a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$.

Since $a > 0$, we can divide the inequality by $a$:

$\frac{-1}{\sqrt{3}} < \frac{x}{a} < \frac{1}{\sqrt{3}}$

Substitute $\frac{x}{a} = \tan\theta$:

$-\frac{1}{\sqrt{3}} < \tan\theta < \frac{1}{\sqrt{3}}$

Taking the inverse tangent, and using the principal value range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$:

$\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) < \theta < \tan^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$-\frac{\pi}{6} < \theta < \frac{\pi}{6}$

Now substitute $\frac{x}{a} = \tan\theta$ into the simplified expression inside $\tan^{-1}$:

$\frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} = \tan(3\theta)$

So the original function becomes:

$\tan^{-1}(\tan(3\theta))$

We need to check if the argument $3\theta$ lies within the principal value range of $\tan^{-1}$, which is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Since $-\frac{\pi}{6} < \theta < \frac{\pi}{6}$, multiply by 3:

$3 \times \left(-\frac{\pi}{6}\right) < 3\theta < 3 \times \left(\frac{\pi}{6}\right)$

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

The range of $3\theta$ is exactly the principal value range for $\tan^{-1}$.

Therefore, we can simplify $\tan^{-1}(\tan(3\theta)) = 3\theta$.

Finally, substitute back $\theta = \tan^{-1}\left(\frac{x}{a}\right)$.

The simplified expression is $3\theta = 3\tan^{-1}\left(\frac{x}{a}\right)$.

The simplest form of $\tan^{-1}\left( \frac{3a^{2}x-x^{3}}{a^{3}-3ax^{2}} \right)$ is $\mathbf{3\tan^{-1}\left(\frac{x}{a}\right)}$.

We need to find the value of the expression $\tan^{-1} \left[2 \cos\left( 2\sin^{-1} \frac{1}{2}\right) \right]$.

First, let's evaluate the innermost part, $\sin^{-1} \frac{1}{2}$.

We know that the principal value branch of $\sin^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ and $\frac{\pi}{6} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, we have:

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

Now, substitute this value back into the expression:

$\tan^{-1} \left[2 \cos\left( 2 \times \frac{\pi}{6}\right) \right]$

$= \tan^{-1} \left[2 \cos\left( \frac{\pi}{3}\right) \right]$

We know that $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$. Substituting this value:

$= \tan^{-1} \left[2 \times \frac{1}{2}\right]$

$= \tan^{-1} (1)$

We know that the principal value branch of $\tan^{-1}$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Since $\tan\left(\frac{\pi}{4}\right) = 1$ and $\frac{\pi}{4} \in (-\frac{\pi}{2}, \frac{\pi}{2})$, we have:

$\tan^{-1} (1) = \frac{\pi}{4}$

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

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

We use the fundamental identity relating $\tan^{-1} x$ and $\cot^{-1} x$:

$\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$, for all real numbers $x$.

In our expression, we have $x = a$. So, we can substitute:

$\tan^{-1} a + \cot^{-1} a = \frac{\pi}{2}$

Now substitute this result back into the original expression:

$\cot(\tan^{-1} a + \cot^{-1} a) = \cot\left(\frac{\pi}{2}\right)$

We know the value of $\cot\left(\frac{\pi}{2}\right)$:

$\cot\left(\frac{\pi}{2}\right) = \frac{\cos\left(\frac{\pi}{2}\right)}{\sin\left(\frac{\pi}{2}\right)} = \frac{0}{1} = 0$

Therefore, $\cot(\tan^{-1} a + \cot^{-1} a) = \mathbf{0}$.

We need to find the value of the expression $\tan\frac{1}{2}\left[ \sin^{-1} \frac{2x}{1+x^2}+ \cos^{-1}\frac{1-y^2}{1+y^2} \right]$ given the conditions $|x| < 1$, $y > 0$, and $xy < 1$.

We know the following standard inverse trigonometric identities:

1. $\sin^{-1} \frac{2x}{1+x^2} = 2 \tan^{-1} x$, for $|x| \le 1$. Since the condition $|x| < 1$ is given, this identity holds.

2. $\cos^{-1} \frac{1-y^2}{1+y^2} = 2 \tan^{-1} y$, for $y \ge 0$. Since the condition $y > 0$ is given, this identity holds.

Substitute these identities into the given expression:

$\tan\frac{1}{2}\left[ 2 \tan^{-1} x + 2 \tan^{-1} y \right]$

Factor out 2 from the expression inside the square brackets:

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

Simplify by cancelling the $\frac{1}{2}$ and $2$:

$= \tan(\tan^{-1} x + \tan^{-1} y)$

Now, 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 $xy < 1$.

The condition $xy < 1$ is given in the problem, so we can apply this identity.

Substitute this identity back into the expression:

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

Finally, use the property $\tan(\tan^{-1} z) = z$:

$= \frac{x+y}{1-xy}$

Therefore, the value of the given expression is $\mathbf{\frac{x+y}{1-xy}}$.

Given:

The equation is given as:

$\sin \left( \sin^{-1}\frac{1}{5} + \cos^{-1}x \right) = 1$

To Find:

The value of $x$.

Solution:

We know that $\sin \theta = 1$ implies $\theta = \frac{\pi}{2} + 2n\pi$ for any integer $n$.

In our case, $\theta = \sin^{-1}\frac{1}{5} + \cos^{-1}x$.

The range of $\sin^{-1} y$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and the range of $\cos^{-1} y$ is $[0, \pi]$.

Since $\frac{1}{5}$ is positive, $0 < \sin^{-1}\frac{1}{5} < \frac{\pi}{2}$.

Also, $0 \le \cos^{-1}x \le \pi$.

Therefore, the sum lies in the interval $(0 + 0, \frac{\pi}{2} + \pi) = (0, \frac{3\pi}{2})$.

The only value in the interval $(0, \frac{3\pi}{2})$ for which $\sin \theta = 1$ is $\theta = \frac{\pi}{2}$.

So, we must have:

$\sin^{-1}\frac{1}{5} + \cos^{-1}x = \frac{\pi}{2}$

We know the identity:

$\sin^{-1}y + \cos^{-1}y = \frac{\pi}{2}$ for all $y \in [-1, 1]$.

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

From our equation, we have:

$\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}\frac{1}{5}$

Comparing this with the identity $\cos^{-1}y = \frac{\pi}{2} - \sin^{-1}y$, we can see that $y$ corresponds to $\frac{1}{5}$.

Thus, $\cos^{-1}x = \cos^{-1}\frac{1}{5}$.

Taking cosine on both sides (or by comparing the arguments), we get:

$x = \frac{1}{5}$

Since $\frac{1}{5}$ is in the domain $[-1, 1]$ for $\cos^{-1}x$ and $\sin^{-1}y$, this solution is valid.

Therefore, the value of $x$ is $\mathbf{\frac{1}{5}}$.

Given:

The equation is given as:

$\tan^{-1} \frac{x - 1}{x - 2} + \tan^{-1} \frac{x + 1}{x + 2} = \frac{\pi}{4}$

To Find:

The value of $x$.

Solution:

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

$\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A+B}{1-AB} \right)$, provided $AB < 1$.

Let $A = \frac{x - 1}{x - 2}$ and $B = \frac{x + 1}{x + 2}$.

Applying the identity to the given equation:

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

Now, let's simplify the numerator and the denominator inside the $\tan^{-1}$ function.

Numerator:

$\frac{x - 1}{x - 2} + \frac{x + 1}{x + 2} = \frac{(x - 1)(x + 2) + (x + 1)(x - 2)}{(x - 2)(x + 2)}$

$= \frac{(x^2 + x - 2) + (x^2 - x - 2)}{x^2 - 4}$

$= \frac{2x^2 - 4}{x^2 - 4}$

Denominator:

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

$= 1 - \frac{x^2 - 1}{x^2 - 4}$

$= \frac{(x^2 - 4) - (x^2 - 1)}{x^2 - 4}$

$= \frac{x^2 - 4 - x^2 + 1}{x^2 - 4}$

$= \frac{-3}{x^2 - 4}$

Substitute these simplified parts back into the equation (assuming $x^2 - 4 \neq 0$, i.e., $x \neq \pm 2$):

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

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

Taking the tangent of both sides:

$\frac{2x^2 - 4}{-3} = \tan\left(\frac{\pi}{4}\right)$

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

Multiply both sides by -3:

$2x^2 - 4 = -3$

$2x^2 = -3 + 4$

$2x^2 = 1$

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

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

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

Now, we must check the condition $AB < 1$ for the identity used.

$AB = \left(\frac{x - 1}{x - 2}\right)\left(\frac{x + 1}{x + 2}\right) = \frac{x^2 - 1}{x^2 - 4}$

Substitute $x^2 = \frac{1}{2}$ into this expression:

$AB = \frac{\frac{1}{2} - 1}{\frac{1}{2} - 4} = \frac{-\frac{1}{2}}{-\frac{7}{2}} = \frac{1}{7}$

Since $AB = \frac{1}{7} < 1$, the use of the identity was valid for the obtained values of $x$.

Therefore, the values of $x$ are $\mathbf{\frac{1}{\sqrt{2}}}$ and $\mathbf{-\frac{1}{\sqrt{2}}}$.

We need to find the value of the expression $\sin^{-1} \left(\sin \frac{2\pi}{3} \right)$.

We know that the principal value branch of the inverse sine function, $\sin^{-1} x$, is $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

The identity $\sin^{-1}(\sin x) = x$ holds only if $x$ belongs to this principal value range.

In this case, the angle is $x = \frac{2\pi}{3}$.

We check if $\frac{2\pi}{3}$ lies within the interval $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

Since $\frac{2\pi}{3}$ is greater than $\frac{\pi}{2}$, it does not lie in the principal value range.

Therefore, we cannot directly write $\sin^{-1} \left(\sin \frac{2\pi}{3} \right) = \frac{2\pi}{3}$.

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

Using the trigonometric identity $\sin(\pi - x) = \sin x$, we can write:

$\sin \frac{2\pi}{3} = \sin\left(\pi - \frac{2\pi}{3}\right)$

$\sin \frac{2\pi}{3} = \sin\left(\frac{3\pi - 2\pi}{3}\right)$

$\sin \frac{2\pi}{3} = \sin\left(\frac{\pi}{3}\right)$

Now, the angle $\frac{\pi}{3}$ lies within the principal value range $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ because $-\frac{\pi}{2} \le \frac{\pi}{3} \le \frac{\pi}{2}$.

So, we can substitute $\sin \frac{\pi}{3}$ for $\sin \frac{2\pi}{3}$ in the original expression:

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

Now, since $\frac{\pi}{3}$ is in the principal value range, we can use the identity $\sin^{-1}(\sin x) = x$:

$\sin^{-1} \left(\sin \frac{\pi}{3} \right) = \frac{\pi}{3}$

Therefore, the value of the expression is $\mathbf{\frac{\pi}{3}}$.

We need to find the value of the expression $\tan^{-1} \left(\tan \frac{3\pi}{4} \right)$.

We know that the principal value branch of the inverse tangent function, $\tan^{-1} x$, is $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$.

The identity $\tan^{-1}(\tan x) = x$ holds only if $x$ belongs to this principal value range.

In this case, the angle is $x = \frac{3\pi}{4}$.

We check if $\frac{3\pi}{4}$ lies within the interval $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$.

Since $\frac{3\pi}{4}$ is greater than $\frac{\pi}{2}$, it does not lie in the principal value range.

Therefore, we cannot directly write $\tan^{-1} \left(\tan \frac{3\pi}{4} \right) = \frac{3\pi}{4}$.

We need to find an angle $\theta \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$ such that $\tan \theta = \tan \frac{3\pi}{4}$.

Using the trigonometric identity $\tan(\pi - x) = -\tan x$ or $\tan(x - \pi) = \tan x$. Let's use the second one which relates to the periodicity of $\tan$.

$\tan \frac{3\pi}{4} = \tan\left(\frac{3\pi}{4} - \pi \right)$

$\tan \frac{3\pi}{4} = \tan\left(\frac{3\pi - 4\pi}{4}\right)$

$\tan \frac{3\pi}{4} = \tan\left(-\frac{\pi}{4}\right)$

Now, the angle $-\frac{\pi}{4}$ lies within the principal value range $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$ because $-\frac{\pi}{2} < -\frac{\pi}{4} < \frac{\pi}{2}$.

So, we can substitute $\tan \left(-\frac{\pi}{4}\right)$ for $\tan \frac{3\pi}{4}$ in the original expression:

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

Now, since $-\frac{\pi}{4}$ is in the principal value range, we can use the identity $\tan^{-1}(\tan x) = x$:

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

Therefore, the value of the expression is $\mathbf{-\frac{\pi}{4}}$.

We need to find the value of the expression $\tan \left(\sin^{-1} \frac{3}{5}+\cot^{-1}\frac{3}{2} \right)$.

To simplify this, we will convert the inverse sine and inverse cotangent terms into inverse tangent terms.

Step 1: Convert $\sin^{-1} \frac{3}{5}$ to $\tan^{-1}$ form.

Let $\alpha = \sin^{-1} \frac{3}{5}$. This implies $\sin \alpha = \frac{3}{5}$.

Since the range of $\sin^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $\sin \alpha$ is positive, $\alpha$ must be in the first quadrant, i.e., $\alpha \in \left(0, \frac{\pi}{2}\right]$.

We can find $\cos \alpha$ using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$.

$\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{25 - 9}{25} = \frac{16}{25}$.

Since $\alpha$ is in the first quadrant, $\cos \alpha$ is positive. So, $\cos \alpha = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

Now, we can find $\tan \alpha$:

$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{3/5}{4/5} = \frac{3}{4}$.

Therefore, $\alpha = \tan^{-1} \frac{3}{4}$. So, $\sin^{-1} \frac{3}{5} = \tan^{-1} \frac{3}{4}$.

Step 2: Convert $\cot^{-1} \frac{3}{2}$ to $\tan^{-1}$ form.

Let $\beta = \cot^{-1} \frac{3}{2}$. This implies $\cot \beta = \frac{3}{2}$.

Since the range of $\cot^{-1}$ is $(0, \pi)$ and $\cot \beta$ is positive, $\beta$ must be in the first quadrant, i.e., $\beta \in \left(0, \frac{\pi}{2}\right)$.

We know that $\tan \beta = \frac{1}{\cot \beta}$.

So, $\tan \beta = \frac{1}{3/2} = \frac{2}{3}$.

Therefore, $\beta = \tan^{-1} \frac{2}{3}$. So, $\cot^{-1} \frac{3}{2} = \tan^{-1} \frac{2}{3}$.

Step 3: Substitute the $\tan^{-1}$ forms back into the original expression.

The expression becomes:

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

Step 4: Apply the sum formula for inverse tangent.

We use the identity $\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A+B}{1-AB} \right)$, provided $AB < 1$.

Here $A = \frac{3}{4}$ and $B = \frac{2}{3}$.

Check the condition: $AB = \frac{3}{4} \times \frac{2}{3} = \frac{6}{12} = \frac{1}{2}$. Since $\frac{1}{2} < 1$, the identity is applicable.

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

$= \tan^{-1} \left( \frac{\frac{9 + 8}{12}}{1 - \frac{6}{12}} \right)$

$= \tan^{-1} \left( \frac{\frac{17}{12}}{\frac{12 - 6}{12}} \right)$

$= \tan^{-1} \left( \frac{\frac{17}{12}}{\frac{6}{12}} \right)$

$= \tan^{-1} \left( \frac{17}{\cancel{12}} \times \frac{\cancel{12}}{6} \right)$

$= \tan^{-1} \left( \frac{17}{6} \right)$

Step 5: Substitute the simplified sum back into the expression.

The expression is now:

$\tan \left( \tan^{-1} \frac{17}{6} \right)$

Step 6: Apply the identity $\tan(\tan^{-1} x) = x$.

$\tan \left( \tan^{-1} \frac{17}{6} \right) = \frac{17}{6}$

Therefore, the value of the expression $\tan \left(\sin^{-1} \frac{3}{5}+\cot^{-1}\frac{3}{2} \right)$ is $\mathbf{\frac{17}{6}}$.

We need to find the value of the expression $\cos^{-1}\left( \cos\frac{7\pi}{6} \right)$.

We know that the principal value branch of the inverse cosine function, $\cos^{-1} x$, is $[0, \pi]$.

The identity $\cos^{-1}(\cos x) = x$ holds only if $x$ belongs to this principal value range $[0, \pi]$.

In this case, the angle is $x = \frac{7\pi}{6}$.

We check if $\frac{7\pi}{6}$ lies within the interval $[0, \pi]$.

Since $\frac{7\pi}{6} > \pi$, it does not lie in the principal value range.

Therefore, we cannot directly write $\cos^{-1}\left( \cos\frac{7\pi}{6} \right) = \frac{7\pi}{6}$.

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

Using the trigonometric identity $\cos(2\pi - x) = \cos x$, we can write:

$\cos \frac{7\pi}{6} = \cos\left(2\pi - \frac{7\pi}{6}\right)$

$\cos \frac{7\pi}{6} = \cos\left(\frac{12\pi - 7\pi}{6}\right)$

$\cos \frac{7\pi}{6} = \cos\left(\frac{5\pi}{6}\right)$

Now, the angle $\frac{5\pi}{6}$ lies within the principal value range $[0, \pi]$ because $0 \le \frac{5\pi}{6} \le \pi$.

So, we can substitute $\cos \frac{5\pi}{6}$ for $\cos \frac{7\pi}{6}$ in the original expression:

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

Now, since $\frac{5\pi}{6}$ is in the principal value range, we can use the identity $\cos^{-1}(\cos x) = x$:

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

Therefore, the value of the expression is $\frac{5\pi}{6}$.

Comparing this with the given options, the correct option is (B).

We need to find the value of the expression $\sin \left( \frac{\pi}{3}-\sin^{-1} \left( -\frac{1}{2} \right)\right)$.

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

We know that the principal value branch of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

We also know the identity $\sin^{-1}(-x) = -\sin^{-1} x$.

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

We know that $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ and $\frac{\pi}{6}$ lies in the principal value range $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

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

Substituting this back, we get:

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

Now, substitute this value into the original expression:

$\sin \left( \frac{\pi}{3} - \left(-\frac{\pi}{6}\right) \right)$

$= \sin \left( \frac{\pi}{3} + \frac{\pi}{6} \right)$

To add the angles, find a common denominator:

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

$= \sin \left( \frac{3\pi}{6} \right)$

$= \sin \left( \frac{\pi}{2} \right)$

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

Therefore, the value of the expression is 1.

Comparing this with the given options, the correct option is (D).

We need to find the value of the expression $\tan^{-1} \sqrt{3} - \cot^{-1} (-\sqrt{3})$.

Let's evaluate each term separately.

First term: $\tan^{-1} \sqrt{3}$

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

We are looking for an angle $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$ such that $\tan \theta = \sqrt{3}$.

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

Since $\frac{\pi}{3}$ lies in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, we have:

$\tan^{-1} \sqrt{3} = \frac{\pi}{3}$.

Second term: $\cot^{-1} (-\sqrt{3})$

The principal value branch of $\cot^{-1} x$ is $(0, \pi)$.

We are looking for an angle $\phi \in (0, \pi)$ such that $\cot \phi = -\sqrt{3}$.

We use the identity $\cot^{-1}(-x) = \pi - \cot^{-1} x$ for $x > 0$.

Here, $x = \sqrt{3}$. So, $\cot^{-1}(-\sqrt{3}) = \pi - \cot^{-1}(\sqrt{3})$.

Now, we find $\cot^{-1}(\sqrt{3})$. We need an angle $\psi \in (0, \pi)$ such that $\cot \psi = \sqrt{3}$.

We know that $\cot \frac{\pi}{6} = \sqrt{3}$.

Since $\frac{\pi}{6}$ lies in the interval $(0, \pi)$, we have $\cot^{-1}(\sqrt{3}) = \frac{\pi}{6}$.

Substitute this back: $\cot^{-1}(-\sqrt{3}) = \pi - \frac{\pi}{6} = \frac{6\pi - \pi}{6} = \frac{5\pi}{6}$.

Now, calculate the value of the expression:

$\tan^{-1} \sqrt{3} - \cot^{-1} (-\sqrt{3}) = \frac{\pi}{3} - \frac{5\pi}{6}$

To subtract, find a common denominator, which is 6:

$= \frac{2\pi}{6} - \frac{5\pi}{6}$

$= \frac{2\pi - 5\pi}{6}$

$= \frac{-3\pi}{6}$

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

Therefore, the value of the expression is $-\frac{\pi}{2}$.

Comparing this with the given options, the correct option is (B).

We need to find the value of the expression $\sin^{-1} \left(\sin \frac{3\pi}{5} \right)$.

The principal value branch of the inverse sine function, $\sin^{-1} x$, is the interval $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

The identity $\sin^{-1}(\sin x) = x$ is valid only when $x$ lies within this principal value range.

In this problem, the angle is $x = \frac{3\pi}{5}$.

We need to check if $\frac{3\pi}{5}$ lies in the interval $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

Comparing $\frac{3\pi}{5}$ with $\frac{\pi}{2}$, we have $\frac{3\pi}{5} = \frac{6\pi}{10}$ and $\frac{\pi}{2} = \frac{5\pi}{10}$.

Since $\frac{6\pi}{10} > \frac{5\pi}{10}$, we have $\frac{3\pi}{5} > \frac{\pi}{2}$.

Therefore, $\frac{3\pi}{5}$ does not lie in the principal value range $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

We cannot directly use the identity $\sin^{-1}(\sin x) = x$.

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

We use the trigonometric identity $\sin(\pi - x) = \sin x$.

Let $x = \frac{3\pi}{5}$. Then:

$\sin \frac{3\pi}{5} = \sin\left(\pi - \frac{3\pi}{5}\right)$

$\sin \frac{3\pi}{5} = \sin\left(\frac{5\pi - 3\pi}{5}\right)$

$\sin \frac{3\pi}{5} = \sin\left(\frac{2\pi}{5}\right)$

Now, we check if the angle $\frac{2\pi}{5}$ lies in the principal value range $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

Comparing $\frac{2\pi}{5}$ with $\frac{\pi}{2}$ and $-\frac{\pi}{2}$:

$\frac{2\pi}{5} = \frac{4\pi}{10}$, $\frac{\pi}{2} = \frac{5\pi}{10}$, and $-\frac{\pi}{2} = -\frac{5\pi}{10}$.

Since $-\frac{5\pi}{10} \le \frac{4\pi}{10} \le \frac{5\pi}{10}$, we have $-\frac{\pi}{2} \le \frac{2\pi}{5} \le \frac{\pi}{2}$.

So, the angle $\frac{2\pi}{5}$ lies within the principal value range.

Now we substitute $\sin\left(\frac{2\pi}{5}\right)$ back into the original expression:

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

Since $\frac{2\pi}{5}$ is in the principal value range $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$, we can now apply the identity $\sin^{-1}(\sin x) = x$.

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

Therefore, the value of the expression is $\mathbf{\frac{2\pi}{5}}$.

To Prove:

$\sin^{-1} \frac{3}{5} - \sin^{-1} \frac{8}{17} = \cos^{-1} \frac{84}{85}$

Proof:

Let $\alpha = \sin^{-1} \frac{3}{5}$. Then $\sin \alpha = \frac{3}{5}$.

Since the principal value range of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, and $\sin \alpha > 0$, we have $0 < \alpha < \frac{\pi}{2}$.

Now, we find $\cos \alpha$ using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$.

$\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}$.

Since $0 < \alpha < \frac{\pi}{2}$, $\cos \alpha$ must be positive.

$\cos \alpha = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

Let $\beta = \sin^{-1} \frac{8}{17}$. Then $\sin \beta = \frac{8}{17}$.

Since $0 < \frac{8}{17} \le 1$, we have $0 < \beta < \frac{\pi}{2}$.

Now, we find $\cos \beta$ using the identity $\sin^2 \beta + \cos^2 \beta = 1$.

$\cos^2 \beta = 1 - \sin^2 \beta = 1 - \left(\frac{8}{17}\right)^2 = 1 - \frac{64}{289} = \frac{289 - 64}{289} = \frac{225}{289}$.

Since $0 < \beta < \frac{\pi}{2}$, $\cos \beta$ must be positive.

$\cos \beta = \sqrt{\frac{225}{289}} = \frac{15}{17}$.

Now, consider the expression $\cos(\alpha - \beta)$. We use the cosine difference identity:

$\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$

Substitute the values we found:

$\cos(\alpha - \beta) = \left(\frac{4}{5}\right) \left(\frac{15}{17}\right) + \left(\frac{3}{5}\right) \left(\frac{8}{17}\right)$

$\cos(\alpha - \beta) = \frac{60}{85} + \frac{24}{85}$

$\cos(\alpha - \beta) = \frac{60 + 24}{85} = \frac{84}{85}$

This suggests that $\alpha - \beta$ might be equal to $\cos^{-1} \frac{84}{85}$. To confirm this, we need to ensure that $\alpha - \beta$ lies within the principal value range of $\cos^{-1}$, which is $[0, \pi]$.

We know $0 < \alpha < \frac{\pi}{2}$ and $0 < \beta < \frac{\pi}{2}$.

Also, compare $\sin \alpha = \frac{3}{5} = 0.6$ and $\sin \beta = \frac{8}{17} \approx 0.47$.

Since $\sin \alpha > \sin \beta$ and both $\alpha, \beta$ are in the first quadrant, we must have $\alpha > \beta$.

Thus, $\alpha - \beta > 0$.

Since $\alpha < \frac{\pi}{2}$ and $\beta > 0$, we have $\alpha - \beta < \frac{\pi}{2} - 0 = \frac{\pi}{2}$.

So, $0 < \alpha - \beta < \frac{\pi}{2}$.

The interval $(0, \frac{\pi}{2})$ is a subset of the principal value range of $\cos^{-1}$, which is $[0, \pi]$.

Since $\cos(\alpha - \beta) = \frac{84}{85}$ and $0 < \alpha - \beta < \frac{\pi}{2}$, we can conclude that:

$\alpha - \beta = \cos^{-1} \frac{84}{85}$

Substituting back $\alpha = \sin^{-1} \frac{3}{5}$ and $\beta = \sin^{-1} \frac{8}{17}$:

$\sin^{-1} \frac{3}{5} - \sin^{-1} \frac{8}{17} = \cos^{-1} \frac{84}{85}$

Hence Proved.

To Prove:

$\sin^{-1} \frac{12}{13} + \cos^{-1} \frac{4}{5} + \tan^{-1} \frac{63}{16} = \pi$

Proof:

We will evaluate the Left Hand Side (LHS) and show that it equals $\pi$.

LHS = $\sin^{-1} \frac{12}{13} + \cos^{-1} \frac{4}{5} + \tan^{-1} \frac{63}{16}$

First, we convert $\sin^{-1} \frac{12}{13}$ and $\cos^{-1} \frac{4}{5}$ into the form of $\tan^{-1}$.

Convert $\sin^{-1} \frac{12}{13}$ to $\tan^{-1}$ form:

Let $\alpha = \sin^{-1} \frac{12}{13}$. Then $\sin \alpha = \frac{12}{13}$.

Since $\sin \alpha$ is positive, $\alpha$ is in the first quadrant, $0 < \alpha < \frac{\pi}{2}$.

Using $\cos^2 \alpha = 1 - \sin^2 \alpha$:

$\cos^2 \alpha = 1 - \left(\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169}$.

Since $\alpha$ is in the first quadrant, $\cos \alpha > 0$.

$\cos \alpha = \sqrt{\frac{25}{169}} = \frac{5}{13}$.

Now, $\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{12/13}{5/13} = \frac{12}{5}$.

So, $\alpha = \tan^{-1} \frac{12}{5}$. Thus, $\sin^{-1} \frac{12}{13} = \tan^{-1} \frac{12}{5}$.

Convert $\cos^{-1} \frac{4}{5}$ to $\tan^{-1}$ form:

Let $\beta = \cos^{-1} \frac{4}{5}$. Then $\cos \beta = \frac{4}{5}$.

Since $\cos \beta$ is positive, $\beta$ is in the first quadrant, $0 < \beta < \frac{\pi}{2}$.

Using $\sin^2 \beta = 1 - \cos^2 \beta$:

$\sin^2 \beta = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{25 - 16}{25} = \frac{9}{25}$.

Since $\beta$ is in the first quadrant, $\sin \beta > 0$.

$\sin \beta = \sqrt{\frac{9}{25}} = \frac{3}{5}$.

Now, $\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{3/5}{4/5} = \frac{3}{4}$.

So, $\beta = \tan^{-1} \frac{3}{4}$. Thus, $\cos^{-1} \frac{4}{5} = \tan^{-1} \frac{3}{4}$.

Substitute these back into the LHS:

LHS = $\tan^{-1} \frac{12}{5} + \tan^{-1} \frac{3}{4} + \tan^{-1} \frac{63}{16}$

Now, we use the identity for the sum of two inverse tangents. Let $A = \frac{12}{5}$ and $B = \frac{3}{4}$.

Calculate $AB = \frac{12}{5} \times \frac{3}{4} = \frac{36}{20} = \frac{9}{5}$.

Since $A > 0$, $B > 0$ and $AB = \frac{9}{5} > 1$, we use the identity:

$\tan^{-1} A + \tan^{-1} B = \pi + \tan^{-1} \left( \frac{A+B}{1-AB} \right)$

$\tan^{-1} \frac{12}{5} + \tan^{-1} \frac{3}{4} = \pi + \tan^{-1} \left( \frac{\frac{12}{5} + \frac{3}{4}}{1 - \frac{12}{5} \times \frac{3}{4}} \right)$

$= \pi + \tan^{-1} \left( \frac{\frac{48 + 15}{20}}{1 - \frac{36}{20}} \right)$

$= \pi + \tan^{-1} \left( \frac{\frac{63}{20}}{\frac{20 - 36}{20}} \right)$

$= \pi + \tan^{-1} \left( \frac{\frac{63}{20}}{\frac{-16}{20}} \right)$

$= \pi + \tan^{-1} \left( -\frac{63}{16} \right)$

Using the identity $\tan^{-1}(-x) = -\tan^{-1} x$:

$\tan^{-1} \frac{12}{5} + \tan^{-1} \frac{3}{4} = \pi - \tan^{-1} \left( \frac{63}{16} \right)$

Substitute this result back into the expression for LHS:

LHS = $\left( \tan^{-1} \frac{12}{5} + \tan^{-1} \frac{3}{4} \right) + \tan^{-1} \frac{63}{16}$

LHS = $\left( \pi - \tan^{-1} \frac{63}{16} \right) + \tan^{-1} \frac{63}{16}$

LHS = $\pi - \tan^{-1} \frac{63}{16} + \tan^{-1} \frac{63}{16}$

LHS = $\pi$

Since LHS = $\pi$ and RHS = $\pi$, we have shown that the identity holds.

Hence Proved.

Given:

The expression $\tan^{-1}\left[ \frac{a\cos x - b\sin x}{b\cos x + a\sin x} \right]$ and the condition $\frac{a}{b} \tan x > -1$.

To Find:

The simplified form of the given expression.

Solution:

Let the given expression be $E$.

$E = \tan^{-1}\left[ \frac{a\cos x - b\sin x}{b\cos x + a\sin x} \right]$

Assuming $b \cos x \neq 0$, we can divide both the numerator and the denominator inside the $\tan^{-1}$ function by $b \cos x$:

$E = \tan^{-1}\left[ \frac{\frac{a\cos x}{b\cos x} - \frac{b\sin x}{b\cos x}}{\frac{b\cos x}{b\cos x} + \frac{a\sin x}{b\cos x}} \right]$

$E = \tan^{-1}\left[ \frac{\frac{a}{b} - \tan x}{1 + \frac{a}{b} \tan x} \right]$

Let $\frac{a}{b} = \tan y$. Then $y = \tan^{-1} \frac{a}{b}$. Substitute $\tan y$ for $\frac{a}{b}$ in the expression:

$E = \tan^{-1}\left[ \frac{\tan y - \tan x}{1 + \tan y \tan x} \right]$

Using the tangent difference identity, $\tan(y-x) = \frac{\tan y - \tan x}{1 + \tan y \tan x}$, we get:

$E = \tan^{-1}[\tan(y-x)]$

Substitute back $y = \tan^{-1} \frac{a}{b}$:

$E = \tan^{-1}\left[\tan\left(\tan^{-1} \frac{a}{b} - x\right)\right]$

We know that $\tan^{-1}(\tan \theta) = \theta$ if $\theta$ lies in the principal value range $(-\frac{\pi}{2}, \frac{\pi}{2})$.

The given condition $\frac{a}{b} \tan x > -1$ implies $1 + \frac{a}{b} \tan x > 0$.

$1 + \frac{a}{b} \tan x = 1 + \tan y \tan x$. This condition ensures that the denominator in the $\tan(y-x)$ formula is positive.

Also, $1 + \frac{a}{b} \tan x = \frac{b \cos x + a \sin x}{b \cos x}$. So, the condition ensures that $b \cos x + a \sin x$ and $b \cos x$ have the same sign and that $b \cos x + a \sin x \neq 0$.

Assuming that the angle $\left(\tan^{-1} \frac{a}{b} - x\right)$ lies within the principal value range $(-\frac{\pi}{2}, \frac{\pi}{2})$, we can simplify the expression as:

$E = \tan^{-1} \frac{a}{b} - x$

This is the simplest form under the assumption on the range of $\tan^{-1} \frac{a}{b} - x$.

Alternate Solution:

We start from the expression obtained after dividing by $b \cos x$:

$E = \tan^{-1}\left[ \frac{\frac{a}{b} - \tan x}{1 + \frac{a}{b} \tan x} \right]$

Let $A = \frac{a}{b}$ and $B = \tan x$. The expression is $E = \tan^{-1}\left[ \frac{A - B}{1 + AB} \right]$.

We use the identity: $\tan^{-1} A - \tan^{-1} B = \tan^{-1} \left( \frac{A-B}{1+AB} \right)$, which holds if $AB > -1$.

In our case, $AB = \frac{a}{b} \tan x$. The condition given in the problem is $\frac{a}{b} \tan x > -1$, which is exactly $AB > -1$.

Therefore, we can apply the identity:

$E = \tan^{-1} A - \tan^{-1} B$

$E = \tan^{-1} \left( \frac{a}{b} \right) - \tan^{-1} (\tan x)$

If we assume that $x$ lies in the principal value range of the tangent function, i.e., $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$, then $\tan^{-1}(\tan x) = x$.

Under this assumption, the expression simplifies to:

$E = \tan^{-1} \frac{a}{b} - x$

This matches the result from the first method.

Given:

The equation $\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4}$.

To Find:

The value(s) of $x$ that satisfy the equation.

Solution:

We use the identity for the sum of two inverse tangent functions:

$\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A+B}{1-AB} \right)$, provided that $AB < 1$.

In this case, $A = 2x$ and $B = 3x$. Applying the identity to the given equation:

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

This step is valid only if the condition $AB < 1$ holds, which is $(2x)(3x) < 1$, or $6x^2 < 1$.

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

$\tan^{-1} \left( \frac{5x}{1 - 6x^2} \right) = \frac{\pi}{4}$

Taking the tangent of both sides of the equation:

$\frac{5x}{1 - 6x^2} = \tan\left(\frac{\pi}{4}\right)$

Since $\tan\left(\frac{\pi}{4}\right) = 1$, we have:

$\frac{5x}{1 - 6x^2} = 1$

Multiply both sides by $(1 - 6x^2)$, assuming $1 - 6x^2 \neq 0$:

$5x = 1 - 6x^2$

Rearrange the terms to form a quadratic equation:

$6x^2 + 5x - 1 = 0$

We can solve this quadratic equation by factorization. We need two numbers that multiply to $(6)(-1) = -6$ and add up to $5$. These numbers are $6$ and $-1$.

$6x^2 + 6x - x - 1 = 0$

Factor by grouping:

$6x(x + 1) - 1(x + 1) = 0$

$(6x - 1)(x + 1) = 0$

This gives two possible solutions for $x$:

$6x - 1 = 0 \implies x = \frac{1}{6}$

$x + 1 = 0 \implies x = -1$

Now, we must check if these solutions satisfy the condition $6x^2 < 1$ required for the application of the $\tan^{-1}$ sum identity.

For $x = \frac{1}{6}$:

$6x^2 = 6 \left(\frac{1}{6}\right)^2 = 6 \left(\frac{1}{36}\right) = \frac{6}{36} = \frac{1}{6}$.

Since $\frac{1}{6} < 1$, the condition $6x^2 < 1$ is satisfied. Thus, $x = \frac{1}{6}$ is a valid solution.

For $x = -1$:

$6x^2 = 6 (-1)^2 = 6(1) = 6$.

Since $6 \not< 1$, the condition $6x^2 < 1$ is not satisfied. Therefore, $x = -1$ is an extraneous solution.

(Alternatively, if $x=-1$, LHS = $\tan^{-1}(-2) + \tan^{-1}(-3) = -(\tan^{-1}2 + \tan^{-1}3)$. Since $\tan^{-1}2 > 0$ and $\tan^{-1}3 > 0$, the LHS is negative. But RHS = $\frac{\pi}{4}$ is positive. So $x=-1$ cannot be a solution.)

The only valid solution to the equation is $x = \frac{1}{6}$.

Therefore, the solution is $\mathbf{x = \frac{1}{6}}$.

We need to find the value of the expression $\cos^{-1}\left( \cos \frac{13\pi}{6} \right)$.

We know that the principal value branch of the inverse cosine function, $\cos^{-1} x$, is the interval $[0, \pi]$.

The identity $\cos^{-1}(\cos x) = x$ is valid only when $x$ lies within this principal value range $[0, \pi]$.

In this problem, the angle is $x = \frac{13\pi}{6}$.

We check if $\frac{13\pi}{6}$ lies within the interval $[0, \pi]$.

We can write $\frac{13\pi}{6}$ as $2\pi + \frac{\pi}{6}$. Since $2\pi + \frac{\pi}{6} > \pi$, the angle $\frac{13\pi}{6}$ does not lie in the principal value range $[0, \pi]$.

Therefore, we cannot directly use the identity $\cos^{-1}(\cos x) = x$.

We need to find an angle $\theta$ within the principal range $[0, \pi]$ such that $\cos \theta = \cos \frac{13\pi}{6}$.

Using the periodicity of the cosine function, $\cos(2n\pi + x) = \cos x$ for any integer $n$.

$\cos \frac{13\pi}{6} = \cos\left(2\pi + \frac{\pi}{6}\right)$

$\cos \frac{13\pi}{6} = \cos\left(\frac{\pi}{6}\right)$

Now, we check if the angle $\frac{\pi}{6}$ lies in the principal value range $[0, \pi]$.

Since $0 \le \frac{\pi}{6} \le \pi$, the angle $\frac{\pi}{6}$ lies within the principal value range.

So, we can substitute $\cos\left(\frac{\pi}{6}\right)$ back into the original expression:

$\cos^{-1}\left( \cos \frac{13\pi}{6} \right) = \cos^{-1}\left( \cos \frac{\pi}{6} \right)$

Since $\frac{\pi}{6}$ is in the principal value range $[0, \pi]$, we can now apply the identity $\cos^{-1}(\cos x) = x$.

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

Therefore, the value of the expression is $\mathbf{\frac{\pi}{6}}$.

We need to find the value of the expression $\tan^{-1}\left( \tan \frac{7\pi}{6} \right)$.

We know that the principal value branch of the inverse tangent function, $\tan^{-1} x$, is the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

The identity $\tan^{-1}(\tan x) = x$ is valid only when $x$ lies within this principal value range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

In this problem, the angle is $x = \frac{7\pi}{6}$.

We check if $\frac{7\pi}{6}$ lies within the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Since $\frac{7\pi}{6} > \frac{\pi}{2}$ (as $\frac{7}{6} > \frac{1}{2}$), the angle $\frac{7\pi}{6}$ does not lie in the principal value range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Therefore, we cannot directly use the identity $\tan^{-1}(\tan x) = x$.

We need to find an angle $\theta$ within the principal range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $\tan \theta = \tan \frac{7\pi}{6}$.

Using the periodicity of the tangent function, $\tan(\pi + x) = \tan x$.

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

$\tan \frac{7\pi}{6} = \tan\left(\pi + \frac{\pi}{6}\right)$

Using the identity, $\tan\left(\pi + \frac{\pi}{6}\right) = \tan\left(\frac{\pi}{6}\right)$.

So, $\tan \frac{7\pi}{6} = \tan\left(\frac{\pi}{6}\right)$.

Now, we check if the angle $\frac{\pi}{6}$ lies in the principal value range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Since $-\frac{\pi}{2} < \frac{\pi}{6} < \frac{\pi}{2}$, the angle $\frac{\pi}{6}$ lies within the principal value range.

So, we can substitute $\tan\left(\frac{\pi}{6}\right)$ back into the original expression:

$\tan^{-1}\left( \tan \frac{7\pi}{6} \right) = \tan^{-1}\left( \tan \frac{\pi}{6} \right)$

Since $\frac{\pi}{6}$ is in the principal value range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, we can now apply the identity $\tan^{-1}(\tan x) = x$.

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

Therefore, the value of the expression is $\mathbf{\frac{\pi}{6}}$.

To Prove:

$2\sin^{-1} \frac{3}{5} = \tan^{-1} \frac{24}{7}$

Proof:

Let the Left Hand Side (LHS) be $2\sin^{-1} \frac{3}{5}$.

First, we convert $\sin^{-1} \frac{3}{5}$ into the form of $\tan^{-1}$.

Let $\alpha = \sin^{-1} \frac{3}{5}$. This implies $\sin \alpha = \frac{3}{5}$.

Since the principal value range of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, and $\sin \alpha = \frac{3}{5}$ is positive, $\alpha$ must lie in the first quadrant, i.e., $0 < \alpha < \frac{\pi}{2}$.

We find $\cos \alpha$ using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$:

$\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{25 - 9}{25} = \frac{16}{25}$.

Since $\alpha$ is in the first quadrant, $\cos \alpha$ is positive.

$\cos \alpha = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

Now, we can find $\tan \alpha$:

$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{3/5}{4/5} = \frac{3}{4}$.

Since $\tan \alpha = \frac{3}{4}$ and $0 < \alpha < \frac{\pi}{2}$, we have $\alpha = \tan^{-1} \frac{3}{4}$.

So, $\sin^{-1} \frac{3}{5} = \tan^{-1} \frac{3}{4}$.

Substitute this back into the LHS:

LHS = $2\sin^{-1} \frac{3}{5} = 2\tan^{-1} \frac{3}{4}$.

Now we use the identity for $2\tan^{-1} x$:

$2\tan^{-1} x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$, provided $|x| < 1$.

Here, $x = \frac{3}{4}$. Since $|\frac{3}{4}| = \frac{3}{4} < 1$, the identity is applicable.

LHS = $2\tan^{-1} \frac{3}{4} = \tan^{-1}\left(\frac{2 \times \frac{3}{4}}{1 - \left(\frac{3}{4}\right)^2}\right)$

$= \tan^{-1}\left(\frac{\frac{6}{4}}{1 - \frac{9}{16}}\right)$

$= \tan^{-1}\left(\frac{\frac{3}{2}}{\frac{16 - 9}{16}}\right)$

$= \tan^{-1}\left(\frac{\frac{3}{2}}{\frac{7}{16}}\right)$

$= \tan^{-1}\left(\frac{3}{2} \times \frac{16}{7}\right)$

$= \tan^{-1}\left(\frac{3 \times \cancel{16}^8}{\cancel{2} \times 7}\right)$

$= \tan^{-1}\left(\frac{24}{7}\right)$

The Right Hand Side (RHS) of the equation to be proved is $\tan^{-1} \frac{24}{7}$.

Since LHS = $\tan^{-1} \frac{24}{7}$ and RHS = $\tan^{-1} \frac{24}{7}$, we have LHS = RHS.

Hence Proved.

To Prove:

$\sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{77}{36}$

Proof:

Let the Left Hand Side (LHS) be $\sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5}$.

We will convert both inverse sine terms to inverse tangent terms.

Convert $\sin^{-1} \frac{8}{17}$ to $\tan^{-1}$ form:

Let $\alpha = \sin^{-1} \frac{8}{17}$. This implies $\sin \alpha = \frac{8}{17}$.

Since $\sin \alpha$ is positive, and the range of $\sin^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, $\alpha$ must be in the first quadrant, i.e., $0 < \alpha < \frac{\pi}{2}$.

Using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$:

$\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \left(\frac{8}{17}\right)^2 = 1 - \frac{64}{289} = \frac{289 - 64}{289} = \frac{225}{289}$.

Since $\alpha$ is in the first quadrant, $\cos \alpha > 0$.

$\cos \alpha = \sqrt{\frac{225}{289}} = \frac{15}{17}$.

Now, $\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{8/17}{15/17} = \frac{8}{15}$.

Therefore, $\sin^{-1} \frac{8}{17} = \tan^{-1} \frac{8}{15}$.

Convert $\sin^{-1} \frac{3}{5}$ to $\tan^{-1}$ form:

Let $\beta = \sin^{-1} \frac{3}{5}$. This implies $\sin \beta = \frac{3}{5}$.

Since $\sin \beta$ is positive, $\beta$ must be in the first quadrant, i.e., $0 < \beta < \frac{\pi}{2}$.

Using the identity $\sin^2 \beta + \cos^2 \beta = 1$:

$\cos^2 \beta = 1 - \sin^2 \beta = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{25 - 9}{25} = \frac{16}{25}$.

Since $\beta$ is in the first quadrant, $\cos \beta > 0$.

$\cos \beta = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

Now, $\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{3/5}{4/5} = \frac{3}{4}$.

Therefore, $\sin^{-1} \frac{3}{5} = \tan^{-1} \frac{3}{4}$.

Substitute the $\tan^{-1}$ forms back into the LHS:

LHS = $\tan^{-1} \frac{8}{15} + \tan^{-1} \frac{3}{4}$.

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

$\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A+B}{1-AB} \right)$, provided $AB < 1$.

Here $A = \frac{8}{15}$ and $B = \frac{3}{4}$.

Check the condition: $AB = \frac{8}{15} \times \frac{3}{4} = \frac{24}{60} = \frac{2}{5}$.

Since $AB = \frac{2}{5} < 1$, the identity is applicable.

LHS = $\tan^{-1} \left( \frac{\frac{8}{15} + \frac{3}{4}}{1 - \frac{8}{15} \times \frac{3}{4}} \right)$

$= \tan^{-1} \left( \frac{\frac{32 + 45}{60}}{1 - \frac{24}{60}} \right)$

$= \tan^{-1} \left( \frac{\frac{77}{60}}{\frac{60 - 24}{60}} \right)$

$= \tan^{-1} \left( \frac{\frac{77}{60}}{\frac{36}{60}} \right)$

$= \tan^{-1} \left( \frac{77}{\cancel{60}} \times \frac{\cancel{60}}{36} \right)$

$= \tan^{-1} \left( \frac{77}{36} \right)$

The Right Hand Side (RHS) of the equation to be proved is $\tan^{-1} \frac{77}{36}$.

Since LHS = $\tan^{-1} \frac{77}{36}$ and RHS = $\tan^{-1} \frac{77}{36}$, we have LHS = RHS.

Hence Proved.

To Prove:

$\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1} \frac{33}{65}$

Proof:

Let the Left Hand Side (LHS) be $\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13}$.

Let $\alpha = \cos^{-1} \frac{4}{5}$. This implies $\cos \alpha = \frac{4}{5}$.

Since the principal value range of $\cos^{-1} x$ is $[0, \pi]$, and $\cos \alpha = \frac{4}{5}$ is positive, $\alpha$ must lie in the first quadrant, i.e., $0 < \alpha < \frac{\pi}{2}$.

Now, we find $\sin \alpha$ using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$:

$\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{25 - 16}{25} = \frac{9}{25}$.

Since $\alpha$ is in the first quadrant, $\sin \alpha$ is positive.

$\sin \alpha = \sqrt{\frac{9}{25}} = \frac{3}{5}$.

Let $\beta = \cos^{-1} \frac{12}{13}$. This implies $\cos \beta = \frac{12}{13}$.

Since $\cos \beta = \frac{12}{13}$ is positive, $\beta$ must also lie in the first quadrant, i.e., $0 < \beta < \frac{\pi}{2}$.

Now, we find $\sin \beta$ using the identity $\sin^2 \beta + \cos^2 \beta = 1$:

$\sin^2 \beta = 1 - \cos^2 \beta = 1 - \left(\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169}$.

Since $\beta$ is in the first quadrant, $\sin \beta$ is positive.

$\sin \beta = \sqrt{\frac{25}{169}} = \frac{5}{13}$.

Now, consider the expression $\cos(\alpha + \beta)$. We use the cosine sum identity:

$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

Substitute the values we found:

$\cos(\alpha + \beta) = \left(\frac{4}{5}\right) \left(\frac{12}{13}\right) - \left(\frac{3}{5}\right) \left(\frac{5}{13}\right)$

$\cos(\alpha + \beta) = \frac{48}{65} - \frac{15}{65}$

$\cos(\alpha + \beta) = \frac{48 - 15}{65} = \frac{33}{65}$

Since $0 < \alpha < \frac{\pi}{2}$ and $0 < \beta < \frac{\pi}{2}$, their sum $\alpha + \beta$ must satisfy $0 < \alpha + \beta < \pi$.

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

Since $\cos(\alpha + \beta) = \frac{33}{65}$ and $0 < \alpha + \beta < \pi$, we can conclude that:

$\alpha + \beta = \cos^{-1} \frac{33}{65}$

Substitute back $\alpha = \cos^{-1} \frac{4}{5}$ and $\beta = \cos^{-1} \frac{12}{13}$:

$\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1} \frac{33}{65}$

This is the expression we wanted to prove.

Hence Proved.

Alternate Method using Identity:

We use the identity: $\cos^{-1} x + \cos^{-1} y = \cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2})$ for $x, y \in [0, 1]$.

Let $x = \frac{4}{5}$ and $y = \frac{12}{13}$. Both are in $[0, 1]$.

LHS = $\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13}$

$= \cos^{-1} \left( \left(\frac{4}{5}\right)\left(\frac{12}{13}\right) - \sqrt{1-\left(\frac{4}{5}\right)^2} \sqrt{1-\left(\frac{12}{13}\right)^2} \right)$

$= \cos^{-1} \left( \frac{48}{65} - \sqrt{1-\frac{16}{25}} \sqrt{1-\frac{144}{169}} \right)$

$= \cos^{-1} \left( \frac{48}{65} - \sqrt{\frac{9}{25}} \sqrt{\frac{25}{169}} \right)$

$= \cos^{-1} \left( \frac{48}{65} - \left(\frac{3}{5}\right) \left(\frac{5}{13}\right) \right)$

$= \cos^{-1} \left( \frac{48}{65} - \frac{15}{65} \right)$

$= \cos^{-1} \left( \frac{33}{65} \right)$

LHS = RHS.

Hence Proved.

To Prove:

$\cos^{-1} \frac{12}{13} + \sin^{-1} \frac{3}{5} = \sin^{-1} \frac{56}{65}$

Proof:

Let the Left Hand Side (LHS) be $\cos^{-1} \frac{12}{13} + \sin^{-1} \frac{3}{5}$.

Let $\alpha = \cos^{-1} \frac{12}{13}$. This implies $\cos \alpha = \frac{12}{13}$.

Since the principal value range of $\cos^{-1} x$ is $[0, \pi]$, and $\cos \alpha = \frac{12}{13}$ is positive, $\alpha$ must lie in the first quadrant, i.e., $0 < \alpha < \frac{\pi}{2}$.

Now, we find $\sin \alpha$ using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$:

$\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \left(\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169}$.

Since $\alpha$ is in the first quadrant, $\sin \alpha$ is positive.

$\sin \alpha = \sqrt{\frac{25}{169}} = \frac{5}{13}$.

Let $\beta = \sin^{-1} \frac{3}{5}$. This implies $\sin \beta = \frac{3}{5}$.

Since the principal value range of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, and $\sin \beta = \frac{3}{5}$ is positive, $\beta$ must lie in the first quadrant, i.e., $0 < \beta < \frac{\pi}{2}$.

Now, we find $\cos \beta$ using the identity $\sin^2 \beta + \cos^2 \beta = 1$:

$\cos^2 \beta = 1 - \sin^2 \beta = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{25 - 9}{25} = \frac{16}{25}$.

Since $\beta$ is in the first quadrant, $\cos \beta$ is positive.

$\cos \beta = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

Now, consider the expression $\sin(\alpha + \beta)$. We use the sine sum identity:

$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$

Substitute the values we found:

$\sin(\alpha + \beta) = \left(\frac{5}{13}\right) \left(\frac{4}{5}\right) + \left(\frac{12}{13}\right) \left(\frac{3}{5}\right)$

$\sin(\alpha + \beta) = \frac{20}{65} + \frac{36}{65}$

$\sin(\alpha + \beta) = \frac{20 + 36}{65} = \frac{56}{65}$

Since $0 < \alpha < \frac{\pi}{2}$ and $0 < \beta < \frac{\pi}{2}$, their sum $\alpha + \beta$ must satisfy $0 < \alpha + \beta < \pi$.

To determine if $\alpha + \beta$ is in the range of $\sin^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, we check the sign of $\cos(\alpha+\beta)$:

$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

$\cos(\alpha + \beta) = \left(\frac{12}{13}\right) \left(\frac{4}{5}\right) - \left(\frac{5}{13}\right) \left(\frac{3}{5}\right)$

$\cos(\alpha + \beta) = \frac{48}{65} - \frac{15}{65} = \frac{33}{65}$

Since $\sin(\alpha + \beta) > 0$ and $\cos(\alpha + \beta) > 0$, the angle $\alpha + \beta$ must lie in the first quadrant, i.e., $0 < \alpha + \beta < \frac{\pi}{2}$.

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

Since $\sin(\alpha + \beta) = \frac{56}{65}$ and $0 < \alpha + \beta < \frac{\pi}{2}$, we can conclude that:

$\alpha + \beta = \sin^{-1} \frac{56}{65}$

Substitute back $\alpha = \cos^{-1} \frac{12}{13}$ and $\beta = \sin^{-1} \frac{3}{5}$:

$\cos^{-1} \frac{12}{13} + \sin^{-1} \frac{3}{5} = \sin^{-1} \frac{56}{65}$

This is the expression we wanted to prove.

Hence Proved.

To Prove:

$\tan^{-1} \frac{63}{16} = \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}$

Proof:

We will start with the Right Hand Side (RHS) and show that it equals the Left Hand Side (LHS).

RHS = $\sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}$

We convert both terms on the RHS to the $\tan^{-1}$ form.

Convert $\sin^{-1} \frac{5}{13}$ to $\tan^{-1}$ form:

Let $\alpha = \sin^{-1} \frac{5}{13}$. Then $\sin \alpha = \frac{5}{13}$.

Since $\sin \alpha > 0$, $\alpha$ is in the first quadrant ($0 < \alpha < \frac{\pi}{2}$).

Using $\cos^2 \alpha = 1 - \sin^2 \alpha$:

$\cos^2 \alpha = 1 - \left(\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{169 - 25}{169} = \frac{144}{169}$.

Since $\alpha$ is in the first quadrant, $\cos \alpha > 0$.

$\cos \alpha = \sqrt{\frac{144}{169}} = \frac{12}{13}$.

Now, $\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{5/13}{12/13} = \frac{5}{12}$.

So, $\sin^{-1} \frac{5}{13} = \tan^{-1} \frac{5}{12}$.

Convert $\cos^{-1} \frac{3}{5}$ to $\tan^{-1}$ form:

Let $\beta = \cos^{-1} \frac{3}{5}$. Then $\cos \beta = \frac{3}{5}$.

Since $\cos \beta > 0$, $\beta$ is in the first quadrant ($0 < \beta < \frac{\pi}{2}$).

Using $\sin^2 \beta = 1 - \cos^2 \beta$:

$\sin^2 \beta = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{25 - 9}{25} = \frac{16}{25}$.

Since $\beta$ is in the first quadrant, $\sin \beta > 0$.

$\sin \beta = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

Now, $\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{4/5}{3/5} = \frac{4}{3}$.

So, $\cos^{-1} \frac{3}{5} = \tan^{-1} \frac{4}{3}$.

Substitute these $\tan^{-1}$ forms back into the RHS:

RHS = $\tan^{-1} \frac{5}{12} + \tan^{-1} \frac{4}{3}$

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

$\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A+B}{1-AB} \right)$, provided $A > 0, B > 0, AB < 1$.

Let $A = \frac{5}{12}$ and $B = \frac{4}{3}$. Both are positive.

Check the condition $AB$:

$AB = \frac{5}{12} \times \frac{4}{3} = \frac{20}{36} = \frac{5}{9}$.

Since $AB = \frac{5}{9} < 1$, the identity is applicable.

RHS = $\tan^{-1} \left( \frac{\frac{5}{12} + \frac{4}{3}}{1 - \frac{5}{12} \times \frac{4}{3}} \right)$

$= \tan^{-1} \left( \frac{\frac{5 + 16}{12}}{1 - \frac{20}{36}} \right)$

$= \tan^{-1} \left( \frac{\frac{21}{12}}{\frac{36 - 20}{36}} \right)$

$= \tan^{-1} \left( \frac{\frac{21}{12}}{\frac{16}{36}} \right)$

$= \tan^{-1} \left( \frac{21}{12} \times \frac{36}{16} \right)$

$= \tan^{-1} \left( \frac{21 \times \cancel{36}^3}{\cancel{12} \times 16} \right)$

$= \tan^{-1} \left( \frac{21 \times 3}{16} \right)$

$= \tan^{-1} \left( \frac{63}{16} \right)$

The Left Hand Side (LHS) is $\tan^{-1} \frac{63}{16}$.

Since RHS = $\tan^{-1} \frac{63}{16}$ and LHS = $\tan^{-1} \frac{63}{16}$, we have RHS = LHS.

Hence Proved.

To Prove:

$\tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{7} + \tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{8} = \frac{\pi}{4}$

Proof:

Let the Left Hand Side (LHS) be $\tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{7} + \tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{8}$.

We will group the terms and apply the identity $\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A+B}{1-AB} \right)$, provided $AB < 1$.

Consider the first two terms: $\tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{7}$.

Here $A = \frac{1}{5}$ and $B = \frac{1}{7}$. $AB = \frac{1}{5} \times \frac{1}{7} = \frac{1}{35}$. Since $\frac{1}{35} < 1$, the identity is applicable.

$\tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{7} = \tan^{-1} \left( \frac{\frac{1}{5} + \frac{1}{7}}{1 - \frac{1}{5} \times \frac{1}{7}} \right)$

$= \tan^{-1} \left( \frac{\frac{7+5}{35}}{1 - \frac{1}{35}} \right)$

$= \tan^{-1} \left( \frac{\frac{12}{35}}{\frac{35-1}{35}} \right)$

$= \tan^{-1} \left( \frac{\frac{12}{35}}{\frac{34}{35}} \right)$

$= \tan^{-1} \left( \frac{12}{34} \right) = \tan^{-1} \left( \frac{6}{17} \right)$

Consider the last two terms: $\tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{8}$.

Here $A = \frac{1}{3}$ and $B = \frac{1}{8}$. $AB = \frac{1}{3} \times \frac{1}{8} = \frac{1}{24}$. Since $\frac{1}{24} < 1$, the identity is applicable.

$\tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{8} = \tan^{-1} \left( \frac{\frac{1}{3} + \frac{1}{8}}{1 - \frac{1}{3} \times \frac{1}{8}} \right)$

$= \tan^{-1} \left( \frac{\frac{8+3}{24}}{1 - \frac{1}{24}} \right)$

$= \tan^{-1} \left( \frac{\frac{11}{24}}{\frac{24-1}{24}} \right)$

$= \tan^{-1} \left( \frac{\frac{11}{24}}{\frac{23}{24}} \right)$

$= \tan^{-1} \left( \frac{11}{23} \right)$

Now substitute these results back into the LHS:

LHS = $\left( \tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{7} \right) + \left( \tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{8} \right)$

LHS = $\tan^{-1} \frac{6}{17} + \tan^{-1} \frac{11}{23}$

Apply the sum identity again. Here $A = \frac{6}{17}$ and $B = \frac{11}{23}$.

Check the condition $AB$: $AB = \frac{6}{17} \times \frac{11}{23} = \frac{66}{391}$. Since $66 < 391$, $AB < 1$, the identity is applicable.

LHS = $\tan^{-1} \left( \frac{\frac{6}{17} + \frac{11}{23}}{1 - \frac{6}{17} \times \frac{11}{23}} \right)$

$= \tan^{-1} \left( \frac{\frac{6 \times 23 + 11 \times 17}{17 \times 23}}{1 - \frac{66}{391}} \right)$

$= \tan^{-1} \left( \frac{\frac{138 + 187}{391}}{\frac{391 - 66}{391}} \right)$

$= \tan^{-1} \left( \frac{\frac{325}{391}}{\frac{325}{391}} \right)$

$= \tan^{-1} (1)$

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

LHS = $\frac{\pi}{4}$

The Right Hand Side (RHS) is $\frac{\pi}{4}$.

Since LHS = RHS, the identity is proved.

Hence Proved.

To Prove:

$\tan^{-1} \sqrt{x} = \frac{1}{2} \cos^{-1} \left( \frac{1-x}{1+x} \right)$ for $x \in [0, 1]$.

Proof:

Let $\tan^{-1} \sqrt{x} = y$.

Since $x \in [0, 1]$, we have $\sqrt{x} \in [0, 1]$.

The range of $\tan^{-1} u$ for $u \in [0, 1]$ is $[0, \frac{\pi}{4}]$.

So, $y = \tan^{-1} \sqrt{x} \in [0, \frac{\pi}{4}]$.

From $\tan^{-1} \sqrt{x} = y$, we have $\tan y = \sqrt{x}$.

Squaring both sides, we get $\tan^2 y = x$.

Now, consider the Right Hand Side (RHS) of the equation we want to prove:

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

Substitute $x = \tan^2 y$ into the expression inside $\cos^{-1}$:

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

Using the trigonometric identity $\cos(2y) = \frac{1-\tan^2 y}{1+\tan^2 y}$, we have:

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

Substitute this back into the RHS expression:

RHS = $\frac{1}{2} \cos^{-1} (\cos(2y))$

We know that $\cos^{-1}(\cos \theta) = \theta$ if $\theta$ lies in the principal value range $[0, \pi]$.

We need to check the range of $2y$.

Since $y \in [0, \frac{\pi}{4}]$, multiplying by 2 gives $2y \in [0, \frac{\pi}{2}]$.

The interval $[0, \frac{\pi}{2}]$ is a subset of the principal value range $[0, \pi]$.

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

Substituting this into the expression for RHS:

RHS = $\frac{1}{2} (2y) = y$.

We started by letting $y = \tan^{-1} \sqrt{x}$, which is the Left Hand Side (LHS).

So, we have shown that RHS = $y$ and LHS = $y$.

Therefore, LHS = RHS, i.e.,

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

Hence Proved.

To Prove:

$\cot^{-1} \left( \frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}} \right) = \frac{x}{2}$ for $x \in \left( 0, \frac{\pi}{4} \right)$.

Proof:

Let the Left Hand Side (LHS) be $\cot^{-1} \left( \frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}} \right)$.

We first simplify the terms under the square root using the identities:

$1 = \cos^2(x/2) + \sin^2(x/2)$

$\sin x = 2 \sin(x/2) \cos(x/2)$

So, $1 + \sin x = \cos^2(x/2) + \sin^2(x/2) + 2 \sin(x/2) \cos(x/2) = (\cos(x/2) + \sin(x/2))^2$.

And, $1 - \sin x = \cos^2(x/2) + \sin^2(x/2) - 2 \sin(x/2) \cos(x/2) = (\cos(x/2) - \sin(x/2))^2$.

Now, consider the square roots:

$\sqrt{1+\sin x} = \sqrt{(\cos(x/2) + \sin(x/2))^2} = |\cos(x/2) + \sin(x/2)|$.

$\sqrt{1-\sin x} = \sqrt{(\cos(x/2) - \sin(x/2))^2} = |\cos(x/2) - \sin(x/2)|$.

We are given that $x \in \left( 0, \frac{\pi}{4} \right)$. This implies $\frac{x}{2} \in \left( 0, \frac{\pi}{8} \right)$.

For $\frac{x}{2} \in \left( 0, \frac{\pi}{8} \right)$:

Both $\cos(x/2)$ and $\sin(x/2)$ are positive. Thus, $\cos(x/2) + \sin(x/2) > 0$.

Also, in the interval $(0, \frac{\pi}{4})$, the cosine function is greater than the sine function. Since $\frac{x}{2} \in (0, \frac{\pi}{8}) \subset (0, \frac{\pi}{4})$, we have $\cos(x/2) > \sin(x/2)$.

Therefore, $\cos(x/2) - \sin(x/2) > 0$.

So we can remove the absolute value signs:

$\sqrt{1+\sin x} = \cos(x/2) + \sin(x/2)$.

$\sqrt{1-\sin x} = \cos(x/2) - \sin(x/2)$.

Substitute these simplified expressions into the fraction inside $\cot^{-1}$:

Numerator = $\sqrt{1+\sin x} + \sqrt{1-\sin x} = (\cos(x/2) + \sin(x/2)) + (\cos(x/2) - \sin(x/2)) = 2\cos(x/2)$.

Denominator = $\sqrt{1+\sin x} - \sqrt{1-\sin x} = (\cos(x/2) + \sin(x/2)) - (\cos(x/2) - \sin(x/2)) = 2\sin(x/2)$.

The fraction becomes $\frac{2\cos(x/2)}{2\sin(x/2)} = \frac{\cos(x/2)}{\sin(x/2)} = \cot(x/2)$.

So, the LHS simplifies to:

LHS = $\cot^{-1}(\cot(x/2))$.

We know that $\cot^{-1}(\cot y) = y$ if $y$ lies in the principal value range of $\cot^{-1}$, which is $(0, \pi)$.

We found that $\frac{x}{2} \in \left( 0, \frac{\pi}{8} \right)$.

Since $\left( 0, \frac{\pi}{8} \right)$ is a subset of $(0, \pi)$, the identity $\cot^{-1}(\cot(x/2)) = x/2$ is applicable.

Therefore, LHS = $\frac{x}{2}$.

The Right Hand Side (RHS) is $\frac{x}{2}$.

Since LHS = RHS, the identity is proved.

Hence Proved.

To Prove:

$\tan^{-1} \left( \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \right) = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x$ for $x \in \left[-\frac{1}{\sqrt{2}}, 1\right]$.

Proof:

Let the Left Hand Side (LHS) be $\tan^{-1} \left( \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \right)$.

Following the hint, let $x = \cos 2\theta$.

First, determine the range for $2\theta$. Since $x \in \left[-\frac{1}{\sqrt{2}}, 1\right]$, we have:

$-\frac{1}{\sqrt{2}} \le \cos 2\theta \le 1$.

The principal value range for $\cos^{-1}$ is $[0, \pi]$. Within this range:

$\cos(0) = 1$

$\cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}$

So, the inequality implies $0 \le 2\theta \le \frac{3\pi}{4}$.

Dividing by 2 gives the range for $\theta$: $0 \le \theta \le \frac{3\pi}{8}$.

Now, substitute $x = \cos 2\theta$ into the LHS expression.

We use the identities $1 + \cos 2\theta = 2 \cos^2 \theta$ and $1 - \cos 2\theta = 2 \sin^2 \theta$.

$\sqrt{1+x} = \sqrt{1+\cos 2\theta} = \sqrt{2 \cos^2 \theta} = \sqrt{2} |\cos \theta|$.

$\sqrt{1-x} = \sqrt{1-\cos 2\theta} = \sqrt{2 \sin^2 \theta} = \sqrt{2} |\sin \theta|$.

Since $\theta \in \left[0, \frac{3\pi}{8}\right]$, which is in the first quadrant (as $\frac{3\pi}{8} < \frac{\pi}{2}$), both $\cos \theta$ and $\sin \theta$ are non-negative.

Therefore, $|\cos \theta| = \cos \theta$ and $|\sin \theta| = \sin \theta$.

$\sqrt{1+x} = \sqrt{2} \cos \theta$.

$\sqrt{1-x} = \sqrt{2} \sin \theta$.

Substitute these into the fraction inside $\tan^{-1}$:

$\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} = \frac{\sqrt{2} \cos \theta - \sqrt{2} \sin \theta}{\sqrt{2} \cos \theta + \sqrt{2} \sin \theta}$

$= \frac{\sqrt{2}(\cos \theta - \sin \theta)}{\sqrt{2}(\cos \theta + \sin \theta)} = \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}$.

Divide the numerator and denominator by $\cos \theta$ (note that $\cos \theta \neq 0$ since $\theta \le \frac{3\pi}{8} < \frac{\pi}{2}$):

$\frac{\frac{\cos \theta}{\cos \theta} - \frac{\sin \theta}{\cos \theta}}{\frac{\cos \theta}{\cos \theta} + \frac{\sin \theta}{\cos \theta}} = \frac{1 - \tan \theta}{1 + \tan \theta}$.

Using the identity $\tan(\frac{\pi}{4} - \theta) = \frac{\tan(\pi/4) - \tan \theta}{1 + \tan(\pi/4)\tan \theta} = \frac{1 - \tan \theta}{1 + \tan \theta}$.

So the expression inside $\tan^{-1}$ is $\tan(\frac{\pi}{4} - \theta)$.

LHS = $\tan^{-1}\left( \tan(\frac{\pi}{4} - \theta) \right)$.

Now we check the range of the angle $\frac{\pi}{4} - \theta$.

Since $0 \le \theta \le \frac{3\pi}{8}$, we have $-\frac{3\pi}{8} \le -\theta \le 0$.

Adding $\frac{\pi}{4}$ throughout: $\frac{\pi}{4} - \frac{3\pi}{8} \le \frac{\pi}{4} - \theta \le \frac{\pi}{4} + 0$.

$\frac{2\pi - 3\pi}{8} \le \frac{\pi}{4} - \theta \le \frac{\pi}{4}$.

$-\frac{\pi}{8} \le \frac{\pi}{4} - \theta \le \frac{\pi}{4}$.

The principal value range for $\tan^{-1}$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$. Since $[-\frac{\pi}{8}, \frac{\pi}{4}] \subset (-\frac{\pi}{2}, \frac{\pi}{2})$, we can apply the identity $\tan^{-1}(\tan y) = y$.

LHS = $\frac{\pi}{4} - \theta$.

Finally, we express $\theta$ in terms of $x$. From $x = \cos 2\theta$, we take $\cos^{-1}$ of both sides:

$\cos^{-1} x = \cos^{-1}(\cos 2\theta)$.

Since $2\theta \in [0, \frac{3\pi}{4}]$, which is within the principal range $[0, \pi]$ for $\cos^{-1}$, we have:

$\cos^{-1} x = 2\theta$.

So, $\theta = \frac{1}{2} \cos^{-1} x$.

Substitute this back into the expression for LHS:

LHS = $\frac{\pi}{4} - \frac{1}{2} \cos^{-1} x$.

This is equal to the Right Hand Side (RHS).

Hence Proved.

To Prove:

$\frac{9\pi}{8} - \frac{9}{4} \sin^{-1} \frac{1}{3} = \frac{9}{4} \sin^{-1} \frac{2\sqrt{2}}{3}$

Proof:

We start by rearranging the equation to be proved:

$\frac{9\pi}{8} = \frac{9}{4} \sin^{-1} \frac{1}{3} + \frac{9}{4} \sin^{-1} \frac{2\sqrt{2}}{3}$

Factor out $\frac{9}{4}$ from the Right Hand Side (RHS):

$\frac{9\pi}{8} = \frac{9}{4} \left( \sin^{-1} \frac{1}{3} + \sin^{-1} \frac{2\sqrt{2}}{3} \right)$

Divide both sides by $\frac{9}{4}$:

$\frac{9\pi}{8} \times \frac{4}{9} = \sin^{-1} \frac{1}{3} + \sin^{-1} \frac{2\sqrt{2}}{3}$

$\frac{\cancel{9}\pi}{\cancel{8}_2} \times \frac{\cancel{4}}{\cancel{9}} = \sin^{-1} \frac{1}{3} + \sin^{-1} \frac{2\sqrt{2}}{3}$

$\frac{\pi}{2} = \sin^{-1} \frac{1}{3} + \sin^{-1} \frac{2\sqrt{2}}{3}$

Now, we need to prove this simplified identity: $\sin^{-1} \frac{1}{3} + \sin^{-1} \frac{2\sqrt{2}}{3} = \frac{\pi}{2}$.

This is equivalent to showing that $\sin^{-1} \frac{2\sqrt{2}}{3} = \frac{\pi}{2} - \sin^{-1} \frac{1}{3}$.

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

Using this identity with $x = \frac{1}{3}$ (which is in $[-1, 1]$), we have:

$\cos^{-1} \frac{1}{3} = \frac{\pi}{2} - \sin^{-1} \frac{1}{3}$.

So, we need to prove that $\sin^{-1} \frac{2\sqrt{2}}{3} = \cos^{-1} \frac{1}{3}$.

Let $\alpha = \cos^{-1} \frac{1}{3}$. This implies $\cos \alpha = \frac{1}{3}$.

Since the range of $\cos^{-1}$ is $[0, \pi]$ and $\cos \alpha > 0$, $\alpha$ must be in the interval $[0, \frac{\pi}{2})$.

Now, find $\sin \alpha$ using $\sin^2 \alpha + \cos^2 \alpha = 1$:

$\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \left(\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9}$.

Since $\alpha \in [0, \frac{\pi}{2})$, $\sin \alpha \ge 0$.

$\sin \alpha = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}$.

From $\sin \alpha = \frac{2\sqrt{2}}{3}$, and knowing that $\alpha \in [0, \frac{\pi}{2})$ (which is within the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ for $\sin^{-1}$), we can write:

$\alpha = \sin^{-1} \frac{2\sqrt{2}}{3}$.

So, we have shown that $\cos^{-1} \frac{1}{3} = \sin^{-1} \frac{2\sqrt{2}}{3}$.

Substituting this back into our requirement:

$\frac{\pi}{2} - \sin^{-1} \frac{1}{3} = \cos^{-1} \frac{1}{3} = \sin^{-1} \frac{2\sqrt{2}}{3}$.

Thus, $\sin^{-1} \frac{1}{3} + \sin^{-1} \frac{2\sqrt{2}}{3} = \frac{\pi}{2}$ is true.

Since this simplified identity is true, the original equation is also true.

Hence Proved.

Given Equation:

$2\tan^{-1} (\cos x) = \tan^{-1} (2 \text{cosec} x)$

To Find:

The value(s) of $x$ that satisfy the equation.

Solution:

We use the identity $2\tan^{-1} y = \tan^{-1}\left(\frac{2y}{1-y^2}\right)$, which is valid for $|y| < 1$.

Let $y = \cos x$. The identity requires $|\cos x| < 1$. This implies $\cos x \neq 1$ and $\cos x \neq -1$, so $x$ cannot be an integer multiple of $\pi$, i.e., $x \neq n\pi$ for any integer $n$.

Applying the identity to the left side of the given equation:

$\tan^{-1}\left(\frac{2\cos x}{1-\cos^2 x}\right) = \tan^{-1} (2 \text{cosec} x)$

Using the trigonometric identity $1 - \cos^2 x = \sin^2 x$, we get:

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

Since $\tan^{-1} A = \tan^{-1} B$ implies $A = B$, we can equate the arguments:

$\frac{2\cos x}{\sin^2 x} = 2 \text{cosec} x$

We know that $\text{cosec} x = \frac{1}{\sin x}$. This requires $\sin x \neq 0$, which means $x \neq n\pi$. This condition is already covered by $|\cos x| < 1$.

Substitute $\text{cosec} x = \frac{1}{\sin x}$ into the equation:

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

Divide both sides by 2:

$\frac{\cos x}{\sin^2 x} = \frac{1}{\sin x}$

Multiply both sides by $\sin^2 x$ (since $\sin x \neq 0$, $\sin^2 x \neq 0$):

$\cos x = \frac{\sin^2 x}{\sin x}$

$\cos x = \sin x$

To find the values of $x$ for which $\cos x = \sin x$, we can divide by $\cos x$ (assuming $\cos x \neq 0$). If $\cos x = 0$, then $x = \frac{\pi}{2} + k\pi$, and $\sin x = \pm 1$, so $\cos x \neq \sin x$. Thus, $\cos x \neq 0$.

Dividing by $\cos x$ gives:

$\frac{\sin x}{\cos x} = 1$

$\tan x = 1$

The general solution for $\tan x = 1$ is:

$x = n\pi + \frac{\pi}{4}$, where $n$ is any integer.

We must verify that these solutions satisfy the condition $|\cos x| < 1$.

For $x = n\pi + \frac{\pi}{4}$, $\cos x = \cos(n\pi + \frac{\pi}{4})$.

If $n$ is even ($n=2k$), $\cos x = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.

If $n$ is odd ($n=2k+1$), $\cos x = \cos(\pi + \frac{\pi}{4}) = -\cos(\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$.

In both cases, $|\cos x| = \frac{1}{\sqrt{2}}$, which is less than 1. So the condition is satisfied.

Therefore, the general solution to the equation is $\mathbf{x = n\pi + \frac{\pi}{4}}$, where $n$ is an integer.

Often, a specific solution like the principal value is expected. The smallest positive value is when $n=0$, giving $x = \frac{\pi}{4}$.

Given Equation:

$\tan^{-1} \frac{1 - x}{1 + x} = \frac{1}{2} \tan^{-1} x$, with the condition $x > 0$.

To Find:

The value(s) of $x$ that satisfy the equation.

Solution:

We start with the given equation:

$\tan^{-1} \frac{1 - x}{1 + x} = \frac{1}{2} \tan^{-1} x$

We can rewrite the term $\frac{1-x}{1+x}$ using the tangent subtraction formula idea.

Let $1 = \tan A$ and $x = \tan B$. Since $1 = \tan(\frac{\pi}{4})$, we can write:

$\frac{1 - x}{1 + x} = \frac{\tan(\frac{\pi}{4}) - x}{1 + \tan(\frac{\pi}{4}) x}$

We know the identity: $\tan^{-1} A - \tan^{-1} B = \tan^{-1} \left( \frac{A-B}{1+AB} \right)$, which holds if $AB > -1$.

Let $A=1$ and $B=x$. Since $x>0$ is given, $AB = 1 \times x = x > 0 > -1$. Thus, the identity is applicable.

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

Substitute this into the LHS of the given equation:

$\tan^{-1} 1 - \tan^{-1} x = \frac{1}{2} \tan^{-1} x$

We know that $\tan^{-1} 1 = \frac{\pi}{4}$.

$\frac{\pi}{4} - \tan^{-1} x = \frac{1}{2} \tan^{-1} x$

Rearrange the terms to solve for $\tan^{-1} x$:

$\frac{\pi}{4} = \tan^{-1} x + \frac{1}{2} \tan^{-1} x$

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

$\frac{\pi}{4} = \frac{3}{2} \tan^{-1} x$

Multiply both sides by $\frac{2}{3}$:

$\tan^{-1} x = \frac{\pi}{4} \times \frac{2}{3}$

$\tan^{-1} x = \frac{2\pi}{12} = \frac{\pi}{6}$

To find $x$, take the tangent of both sides:

$x = \tan\left(\frac{\pi}{6}\right)$

$x = \frac{1}{\sqrt{3}}$

We must check if this solution satisfies the given condition $x > 0$.

Since $\frac{1}{\sqrt{3}} > 0$, the condition is satisfied.

Therefore, the solution to the equation is $\mathbf{x = \frac{1}{\sqrt{3}}}$.

We need to find the value of the expression $\sin(\tan^{-1} x)$, given $|x| < 1$.

Let $\theta = \tan^{-1} x$. By definition, this means $\tan \theta = x$.

The principal value range for $\tan^{-1} x$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. So, $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

We want to find $\sin(\tan^{-1} x) = \sin \theta$.

We can relate $\sin \theta$ to $\tan \theta$ using trigonometric identities or a right triangle.

Method 1: Using identities

We know that $\text{cosec}^2 \theta = 1 + \cot^2 \theta$.

Since $\cot \theta = \frac{1}{\tan \theta} = \frac{1}{x}$ (assuming $x \neq 0$).

$\text{cosec}^2 \theta = 1 + \left(\frac{1}{x}\right)^2 = 1 + \frac{1}{x^2} = \frac{x^2+1}{x^2}$.

Also, $\text{cosec} \theta = \frac{1}{\sin \theta}$, so $\text{cosec}^2 \theta = \frac{1}{\sin^2 \theta}$.

$\frac{1}{\sin^2 \theta} = \frac{x^2+1}{x^2}$.

$\sin^2 \theta = \frac{x^2}{1+x^2}$.

$\sin \theta = \pm \sqrt{\frac{x^2}{1+x^2}} = \pm \frac{|x|}{\sqrt{1+x^2}}$.

Since $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$, $\sin \theta$ has the same sign as $\theta$. Also, $\tan \theta = x$. If $x > 0$, $\theta \in (0, \frac{\pi}{2})$ and $\sin \theta > 0$. If $x < 0$, $\theta \in (-\frac{\pi}{2}, 0)$ and $\sin \theta < 0$. If $x = 0$, $\theta = 0$ and $\sin \theta = 0$.

Therefore, $\sin \theta$ must have the same sign as $x$. So we can write:

$\sin \theta = \frac{x}{\sqrt{1+x^2}}$.

Method 2: Using a right triangle

Let $\theta = \tan^{-1} x$. Then $\tan \theta = x = \frac{x}{1}$.

Assume $x > 0$. Then $\theta$ is in the first quadrant $(0, \frac{\pi}{2})$. We can draw a right triangle where the angle is $\theta$, the opposite side is $x$, and the adjacent side is $1$.

Opposite = $x$

Adjacent = $1$

Hypotenuse = $\sqrt{(\text{Opposite})^2 + (\text{Adjacent})^2} = \sqrt{x^2 + 1^2} = \sqrt{1+x^2}$.

Now, $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{x}{\sqrt{1+x^2}}$.

If $x < 0$, then $\theta \in (-\frac{\pi}{2}, 0)$. Let $x = -y$ where $y > 0$. Then $\tan \theta = -y$. We can consider a triangle with angle $\phi$ where $\tan \phi = y = \frac{y}{1}$. Then $\sin \phi = \frac{y}{\sqrt{1+y^2}}$. Since $\theta = -\phi$, $\sin \theta = \sin(-\phi) = -\sin \phi = -\frac{y}{\sqrt{1+y^2}} = \frac{-y}{\sqrt{1+(-y)^2}} = \frac{x}{\sqrt{1+x^2}}$.

If $x=0$, $\sin(\tan^{-1} 0) = \sin(0) = 0$, and $\frac{0}{\sqrt{1+0^2}}=0$.

The formula holds for all $x$.

Thus, $\sin(\tan^{-1} x) = \frac{x}{\sqrt{1+x^2}}$.

Comparing this with the given options, the correct option is (D).

Given Equation:

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

To Find:

The value(s) of $x$ that satisfy the equation.

Solution:

First, determine the domain of the equation.

For $\sin^{-1}(1-x)$ to be defined, we must have $-1 \le 1-x \le 1$.

$-1 - 1 \le -x \le 1 - 1 \implies -2 \le -x \le 0 \implies 0 \le x \le 2$.

For $\sin^{-1} x$ to be defined, we must have $-1 \le x \le 1$.

Combining these conditions, the domain for $x$ is $[0, 1]$.

Rearrange the given equation:

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

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

The equation becomes:

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

The range of the inverse sine function $\sin^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Therefore, the value of the left side, $\sin^{-1}(1 - \sin y)$, must lie in $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

This means the right side must also lie in this interval:

$-\frac{\pi}{2} \le \frac{\pi}{2} + 2y \le \frac{\pi}{2}$

Subtract $\frac{\pi}{2}$ from all parts of the inequality:

$-\frac{\pi}{2} - \frac{\pi}{2} \le 2y \le \frac{\pi}{2} - \frac{\pi}{2}$

$-\pi \le 2y \le 0$

Divide by 2:

$-\frac{\pi}{2} \le y \le 0$

We have two conditions for $y$:

1. $y \in [0, \frac{\pi}{2}]$ (from the domain $x \in [0, 1]$)

2. $y \in [-\frac{\pi}{2}, 0]$ (from the range constraint of $\sin^{-1}$)

The only value of $y$ that satisfies both conditions is $y = 0$.

Since $y = \sin^{-1} x$, if $y = 0$, then $\sin^{-1} x = 0$.

$x = \sin(0) = 0$.

We must check if $x=0$ satisfies the original equation:

LHS = $\sin^{-1} (1 - 0) - 2 \sin^{-1} (0)$

LHS = $\sin^{-1} (1) - 2(0)$

LHS = $\frac{\pi}{2} - 0 = \frac{\pi}{2}$.

RHS = $\frac{\pi}{2}$.

Since LHS = RHS, $x = 0$ is the only solution.

Alternate Check (Solving the transformed equation):

From $\sin^{-1} (1 - x) = \frac{\pi}{2} + 2 \sin^{-1} x$, take sine on both sides:

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

$1 - x = \cos(2 \sin^{-1} x)$

Using the identity $\cos(2\theta) = 1 - 2\sin^2\theta$, let $\theta = \sin^{-1} x$. Then $\sin \theta = x$.

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

$1 - x = 1 - 2x^2$

$2x^2 - x = 0$

$x(2x - 1) = 0$

This gives potential solutions $x=0$ and $x=\frac{1}{2}$.

Check $x=0$: $\sin^{-1}(1) - 2\sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$. (Valid)

Check $x=1/2$: $\sin^{-1}(1-\frac{1}{2}) - 2\sin^{-1}(\frac{1}{2}) = \sin^{-1}(\frac{1}{2}) - 2\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6} - 2(\frac{\pi}{6}) = \frac{\pi}{6} - \frac{2\pi}{6} = -\frac{\pi}{6}$.

Since $-\frac{\pi}{6} \neq \frac{\pi}{2}$, $x=\frac{1}{2}$ is an extraneous solution introduced by taking the sine.

The only valid solution is $x=0$.

Comparing this with the given options, the correct option is (C).

Let the given expression be $E$.

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

We use the identity $\tan^{-1} A - \tan^{-1} B = \tan^{-1} \left( \frac{A-B}{1+AB} \right)$, which holds if $AB > -1$.

Let $A = \frac{x}{y}$ and $B = \frac{x-y}{x+y}$.

Calculate the terms needed for the identity:

$A - B = \frac{x}{y} - \frac{x-y}{x+y} = \frac{x(x+y) - y(x-y)}{y(x+y)} = \frac{x^2 + xy - xy + y^2}{y(x+y)} = \frac{x^2 + y^2}{y(x+y)}$

$AB = \left(\frac{x}{y}\right) \left(\frac{x-y}{x+y}\right) = \frac{x(x-y)}{y(x+y)} = \frac{x^2 - xy}{y(x+y)}$

$1 + AB = 1 + \frac{x^2 - xy}{y(x+y)} = \frac{y(x+y) + (x^2 - xy)}{y(x+y)} = \frac{xy + y^2 + x^2 - xy}{y(x+y)} = \frac{x^2 + y^2}{y(x+y)}$

Now compute the fraction inside the resulting $\tan^{-1}$:

$\frac{A-B}{1+AB} = \frac{\frac{x^2 + y^2}{y(x+y)}}{\frac{x^2 + y^2}{y(x+y)}}$

Assuming $x^2 + y^2 \neq 0$ and $y(x+y) \neq 0$, this fraction simplifies to $1$.

Now we need to check the condition $AB > -1$.

$AB > -1 \implies \frac{x^2 - xy}{y(x+y)} > -1$

$\frac{x^2 - xy}{y(x+y)} + 1 > 0$

$\frac{x^2 - xy + y(x+y)}{y(x+y)} > 0$

$\frac{x^2 - xy + xy + y^2}{y(x+y)} > 0$

$\frac{x^2 + y^2}{y(x+y)} > 0$

Since $x^2 + y^2 > 0$ (unless $x=y=0$, which makes the original expression undefined), the condition $AB > -1$ is equivalent to $y(x+y) > 0$.

Case 1: $y(x+y) > 0$ (i.e., $AB > -1$)

In this case, the identity $\tan^{-1} A - \tan^{-1} B = \tan^{-1} \left( \frac{A-B}{1+AB} \right)$ holds.

$E = \tan^{-1}(1)$

The principal value of $\tan^{-1}(1)$ is $\frac{\pi}{4}$.

Case 2: $y(x+y) < 0$ (i.e., $AB < -1$)

In this case, the identity changes based on the signs of $A$ and $B$. If $A = x/y < 0$ and $B = (x-y)/(x+y) > 0$, then $\tan^{-1} A - \tan^{-1} B = -\pi + \tan^{-1} \left( \frac{A-B}{1+AB} \right)$.

This condition ($A<0, B>0, AB<-1$) occurs when $x$ and $y$ have opposite signs and $|x| > |y|$.

In this situation, $E = -\pi + \tan^{-1}(1) = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4}$.

If $A = x/y > 0$ and $B = (x-y)/(x+y) < 0$, then $\tan^{-1} A - \tan^{-1} B = \pi + \tan^{-1} \left( \frac{A-B}{1+AB} \right)$. However, this condition requires $x,y$ to have the same sign ($A>0$) and $y(x+y) < 0$, which is impossible.

So the possible values for the expression are $\frac{\pi}{4}$ and $-\frac{3\pi}{4}$.

Since no conditions on $x$ and $y$ are given, it is standard practice in such multiple-choice questions to assume the conditions for the simplest form of the identity hold ($AB > -1$). Under this assumption, the result is $\frac{\pi}{4}$.

Alternate Method (Substitution):

Let $\frac{x}{y} = \tan \alpha$. Assume $\alpha \in (-\frac{\pi}{2}, \frac{\pi}{2})$.

Then $\tan^{-1}\left(\frac{x}{y}\right) = \alpha$.

The second term is $\tan^{-1} \frac{x-y}{x+y}$. Divide numerator and denominator by $y$ (assuming $y \neq 0$):

$\tan^{-1} \frac{\frac{x}{y} - 1}{\frac{x}{y} + 1} = \tan^{-1} \frac{\tan \alpha - 1}{\tan \alpha + 1}$

$= \tan^{-1} \frac{\tan \alpha - \tan(\pi/4)}{1 + \tan \alpha \tan(\pi/4)}$ (since $\tan(\pi/4)=1$)

$= \tan^{-1}(\tan(\alpha - \frac{\pi}{4}))$

The expression becomes $E = \alpha - \tan^{-1}(\tan(\alpha - \frac{\pi}{4}))$.

If $\alpha - \frac{\pi}{4}$ lies in the principal value range $(-\frac{\pi}{2}, \frac{\pi}{2})$, then $\tan^{-1}(\tan(\alpha - \frac{\pi}{4})) = \alpha - \frac{\pi}{4}$.

This happens when $-\frac{\pi}{2} < \alpha - \frac{\pi}{4} < \frac{\pi}{2}$, which means $-\frac{\pi}{4} < \alpha < \frac{3\pi}{4}$.

Since we initially assumed $\alpha \in (-\frac{\pi}{2}, \frac{\pi}{2})$, this condition is satisfied if $\alpha \in (-\frac{\pi}{4}, \frac{\pi}{2})$.

If $\alpha \in (-\frac{\pi}{4}, \frac{\pi}{2})$, then $E = \alpha - (\alpha - \frac{\pi}{4}) = \frac{\pi}{4}$.

This condition $\alpha \in (-\frac{\pi}{4}, \frac{\pi}{2})$ means $\tan \alpha = x/y > \tan(-\pi/4) = -1$. So $x/y > -1$. This corresponds to $y(x+y)>0$.

If $\alpha - \frac{\pi}{4}$ lies outside $(-\frac{\pi}{2}, \frac{\pi}{2})$, specifically if $\alpha \in (-\frac{\pi}{2}, -\frac{\pi}{4}]$, then $\alpha - \frac{\pi}{4} \in (-\frac{3\pi}{4}, -\frac{\pi}{2}]$. In this case, $\tan^{-1}(\tan(\alpha - \frac{\pi}{4})) = (\alpha - \frac{\pi}{4}) + \pi = \alpha + \frac{3\pi}{4}$.

Then $E = \alpha - (\alpha + \frac{3\pi}{4}) = -\frac{3\pi}{4}$.

This condition $\alpha \in (-\frac{\pi}{2}, -\frac{\pi}{4}]$ means $\tan \alpha = x/y \le -1$. This corresponds to $y(x+y)<0$.

Assuming the standard case where $x/y > -1$, the answer is $\frac{\pi}{4}$.

Comparing with the options, the most likely intended answer is $\mathbf{\frac{\pi}{4}}$, which is option (C).