Chapter 2 Inverse Trigonometric Functions

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

This means $\sin y = \frac{1}{\sqrt{2}}$.

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

So, we have $\sin y = \sin \left(\frac{\pi}{4}\right)$.

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

We need to find a value of $y$ in this interval such that $\sin y = \frac{1}{\sqrt{2}}$.

The angle $\frac{\pi}{4}$ lies in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

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

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

By the definition of inverse trigonometric functions, this means:

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

Since $\cot y$ is negative, $y$ must lie in the second quadrant, as the principal value branch of $\cot^{-1}x$ is $(0, \pi)$.

We use the identity $\cot(\pi - \theta) = -\cot \theta$.

Using this identity in equation (i):

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

We need to find a value of $y$ in this interval such that $\cot y = -\frac{1}{\sqrt{3}}$.

The angle $\frac{2\pi}{3}$ lies in the interval $(0, \pi)$.

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

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

By the definition of inverse trigonometric functions, this means:

$\sin y = -\frac{1}{2}$.

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

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

For $x \in [-1, 1]$, the value of $\sin^{-1}x$ is such that $\sin(\sin^{-1}x) = x$ and $-\frac{\pi}{2} \leq \sin^{-1}x \leq \frac{\pi}{2}$.

Since $\sin y$ is negative, $y$ must be in the interval $[-\frac{\pi}{2}, 0]$.

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

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

$\sin y = \sin\left(-\frac{\pi}{6}\right)$.

The angle $-\frac{\pi}{6}$ lies in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

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

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

By the definition of inverse trigonometric functions, this means:

$\cos y = \frac{\sqrt{3}}{2}$.

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

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

For $x \in [-1, 1]$, the value of $\cos^{-1}x$ is such that $\cos(\cos^{-1}x) = x$ and $0 \leq \cos^{-1}x \leq \pi$.

We need to find a value of $y$ in the interval $[0, \pi]$ such that $\cos y = \frac{\sqrt{3}}{2}$.

The angle $\frac{\pi}{6}$ satisfies $\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$.

The angle $\frac{\pi}{6}$ lies in the interval $[0, \pi]$.

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

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

By the definition of inverse trigonometric functions, this means:

$\text{cosec } y = 2$.

We know that $\text{cosec } y = \frac{1}{\sin y}$.

So, $\frac{1}{\sin y} = 2$, which implies:

$\sin y = \frac{1}{2}$.

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

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

For $x \in \mathbb{R} - (-1, 1)$, the value of $\text{cosec}^{-1}x$ is such that $\text{cosec}(\text{cosec}^{-1}x) = x$ and $-\frac{\pi}{2} \leq \text{cosec}^{-1}x \leq \frac{\pi}{2}$, $\text{cosec}^{-1}x \neq 0$.

We need to find a value of $y$ in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$ such that $\sin y = \frac{1}{2}$.

The angle $\frac{\pi}{6}$ satisfies $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$.

The angle $\frac{\pi}{6}$ lies in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$.

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

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

By the definition of inverse trigonometric functions, this means:

$\tan y = -\sqrt{3}$.

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

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

For $x \in \mathbb{R}$, the value of $\tan^{-1}x$ is such that $\tan(\tan^{-1}x) = x$ and $-\frac{\pi}{2} < \tan^{-1}x < \frac{\pi}{2}$.

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

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

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

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

The angle $-\frac{\pi}{3}$ lies in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.

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

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

By the definition of inverse trigonometric functions, this means:

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

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

For $x \in [-1, 1]$, the value of $\cos^{-1}x$ is such that $\cos(\cos^{-1}x) = x$ and $0 \leq \cos^{-1}x \leq \pi$.

Since $\cos y$ is negative, $y$ must lie in the second quadrant within the principal value branch $[0, \pi]$.

We use the identity $\cos(\pi - \theta) = -\cos \theta$.

Using this identity with the reference angle $\frac{\pi}{3}$ in equation (i):

The angle $\frac{2\pi}{3}$ lies in the principal value branch $[0, \pi]$.

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

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

By the definition of inverse trigonometric functions, this means:

$\tan y = -1$.

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

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

For $x \in \mathbb{R}$, the value of $\tan^{-1}x$ is such that $\tan(\tan^{-1}x) = x$ and $-\frac{\pi}{2} < \tan^{-1}x < \frac{\pi}{2}$.

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

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

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

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

The angle $-\frac{\pi}{4}$ lies in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.

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

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

By the definition of inverse trigonometric functions, this means:

$\sec y = \frac{2}{\sqrt{3}}$.

We know that $\sec y = \frac{1}{\cos y}$.

So, $\frac{1}{\cos y} = \frac{2}{\sqrt{3}}$, which implies:

$\cos y = \frac{\sqrt{3}}{2}$.

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

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

For $x \in \mathbb{R} - (-1, 1)$, the value of $\sec^{-1}x$ is such that $\sec(\sec^{-1}x) = x$ and $0 \leq \sec^{-1}x \leq \pi$, $\sec^{-1}x \neq \frac{\pi}{2}$.

We need to find a value of $y$ in the interval $[0, \pi] - \{\frac{\pi}{2}\}$ such that $\cos y = \frac{\sqrt{3}}{2}$.

The angle $\frac{\pi}{6}$ satisfies $\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$, and therefore $\sec \left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}$.

The angle $\frac{\pi}{6}$ lies in the interval $[0, \pi] - \{\frac{\pi}{2}\}$.

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

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

By the definition of inverse trigonometric functions, this means:

$\cot y = \sqrt{3}$.

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

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

For $x \in \mathbb{R}$, the value of $\cot^{-1}x$ is such that $\cot(\cot^{-1}x) = x$ and $0 < \cot^{-1}x < \pi$.

We need to find a value of $y$ in the interval $(0, \pi)$ such that $\cot y = \sqrt{3}$.

The angle $\frac{\pi}{6}$ satisfies $\cot \left(\frac{\pi}{6}\right) = \sqrt{3}$.

The angle $\frac{\pi}{6}$ lies in the interval $(0, \pi)$.

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

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

By the definition of inverse trigonometric functions, this means:

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

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

For $x \in [-1, 1]$, the value of $\cos^{-1}x$ is such that $\cos(\cos^{-1}x) = x$ and $0 \leq \cos^{-1}x \leq \pi$.

Since $\cos y$ is negative, $y$ must lie in the second quadrant within the principal value branch $[0, \pi]$.

We use the identity $\cos(\pi - \theta) = -\cos \theta$.

Using this identity with the reference angle $\frac{\pi}{4}$ in equation (i):

The angle $\frac{3\pi}{4}$ lies in the principal value branch $[0, \pi]$.

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

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

By the definition of inverse trigonometric functions, this means:

$\text{cosec } y = -\sqrt{2}$.

We know that $\text{cosec } y = \frac{1}{\sin y}$.

So, $\frac{1}{\sin y} = -\sqrt{2}$, which implies:

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

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

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

For $x \in \mathbb{R} - (-1, 1)$, the value of $\text{cosec}^{-1}x$ is such that $\text{cosec}(\text{cosec}^{-1}x) = x$ and $-\frac{\pi}{2} \leq \text{cosec}^{-1}x \leq \frac{\pi}{2}$, $\text{cosec}^{-1}x \neq 0$.

We need to find a value of $y$ in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$ such that $\sin y = -\frac{1}{\sqrt{2}}$.

Since $\sin y$ is negative, $y$ must be in the interval $[-\frac{\pi}{2}, 0)$.

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

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

$\sin y = \sin\left(-\frac{\pi}{4}\right)$.

The angle $-\frac{\pi}{4}$ lies in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$.

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

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

We will find the principal value of each term separately.

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

This means $\tan y_1 = 1$.

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

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

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

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

This means $\cos y_2 = -\frac{1}{2}$.

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, $y_2$ is in the second quadrant within the principal value branch.

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

$\cos y_2 = -\cos\left(\frac{\pi}{3}\right) = \cos\left(\pi - \frac{\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right)$.

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

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

This means $\sin y_3 = -\frac{1}{2}$.

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

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

Since $\sin y_3$ is negative, $y_3$ is in the interval $[-\frac{\pi}{2}, 0]$.

Using the identity $\sin(-\theta) = -\sin \theta$:

$\sin y_3 = -\sin\left(\frac{\pi}{6}\right) = \sin\left(-\frac{\pi}{6}\right)$.

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

Now, we add the principal values from (i), (ii), and (iii):

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

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

To add these fractions, find a common denominator, which is the LCM of 4, 3, and 6. The LCM is 12.

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

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

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

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

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

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 3.

$= \frac{\cancel{9}^{3}\pi}{\cancel{12}_{4}}$

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

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

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

We will find the principal value of each inverse trigonometric function separately.

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

This means $\cos y_1 = \frac{1}{2}$.

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

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

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

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

This means $\sin y_2 = \frac{1}{2}$.

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

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

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

Now, we substitute the principal values from (i) and (ii) into the given expression:

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

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

Simplify the second term:

$= \frac{\pi}{3} + \frac{\cancel{2}^{1}\pi}{\cancel{6}_{3}}$

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

Add the terms:

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

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

The value of the expression is $\frac{2\pi}{3}$.

We are given that $\sin^{-1} x = y$.

The function $\sin^{-1}x$ is defined as the inverse of the sine function restricted to a specific interval to make it a bijection.

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

This means that for any value of $x$ in the domain of $\sin^{-1}x$ (which is $[-1, 1]$), the principal value $y = \sin^{-1}x$ must lie within the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

So, if $\sin^{-1} x = y$, then $y$ must satisfy the condition:

Comparing this condition with the given options:

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

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

(C) 0 < y < π (This is related to the range for $\cot^{-1}x$).

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

The correct option is (B).

We need to find the value of $\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2)$.

We will find the principal value of each inverse trigonometric function separately.

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

This means $\tan y_1 = \sqrt{3}$.

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

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

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

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

This means $\sec y_2 = -2$.

We know that $\sec y_2 = \frac{1}{\cos y_2}$. So, $\frac{1}{\cos y_2} = -2$, which implies $\cos y_2 = -\frac{1}{2}$.

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

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 principal value branch $[0, \pi]$.

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

$\cos y_2 = -\cos\left(\frac{\pi}{3}\right) = \cos\left(\pi - \frac{\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right)$.

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

Now, we substitute the principal values from (i) and (ii) into the given expression:

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

Combine the terms:

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

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

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

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

Solution:

(i) To Show: $\sin^{-1} \left( 2x\sqrt{1-x^{2}} \right) = 2\sin^{-1} x$ for $-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$.

Let $x = \sin \theta$.

Since $\sin^{-1} x = \theta$, and the principal value branch of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, $\theta$ lies in this interval.

The given condition on $x$ is $-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$.

Substituting $x = \sin \theta$, we get $-\frac{1}{\sqrt{2}} \leq \sin \theta \leq \frac{1}{\sqrt{2}}$.

Since $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, this inequality implies:

Now consider the expression inside $\sin^{-1}$ on the left side:

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

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

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

For $-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$, $\cos \theta$ is non-negative, so $|\cos\theta| = \cos\theta$.

Thus, $2x\sqrt{1-x^{2}} = 2\sin \theta \cos\theta = \sin(2\theta)$.

Now the left side of the identity is $\sin^{-1}(\sin(2\theta))$.

For $\sin^{-1}(\sin \phi) = \phi$ to hold, $\phi$ must be in the principal value branch of $\sin^{-1}$, i.e., $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

In our case, $\phi = 2\theta$. From $-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$, we have:

Since $2\theta$ lies in $[-\frac{\pi}{2}, \frac{\pi}{2}]$, we have $\sin^{-1}(\sin(2\theta)) = 2\theta$.

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

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

This holds for $-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$.

(ii) To Show: $\sin^{-1} \left( 2x\sqrt{1-x^{2}} \right) = 2 \cos^{-1} x$ for $\frac{1}{\sqrt{2}} \leq x \leq 1$.

Let $x = \cos \theta$.

Since $\cos^{-1} x = \theta$, and the principal value branch of $\cos^{-1} x$ is $[0, \pi]$, $\theta$ lies in this interval.

The given condition on $x$ is $\frac{1}{\sqrt{2}} \leq x \leq 1$.

Substituting $x = \cos \theta$, we get $\frac{1}{\sqrt{2}} \leq \cos \theta \leq 1$.

Since $\theta \in [0, \pi]$, and cosine is decreasing on $[0, \pi]$, this inequality implies:

Now consider the expression inside $\sin^{-1}$ on the left side:

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

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

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

For $0 \leq \theta \leq \frac{\pi}{4}$, $\sin \theta$ is non-negative, so $|\sin\theta| = \sin\theta$.

Thus, $2x\sqrt{1-x^{2}} = 2\cos \theta \sin\theta = \sin(2\theta)$.

Now the left side of the identity is $\sin^{-1}(\sin(2\theta))$.

For $\sin^{-1}(\sin \phi) = \phi$ to hold, $\phi$ must be in the principal value branch of $\sin^{-1}$, i.e., $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

In our case, $\phi = 2\theta$. From $0 \leq \theta \leq \frac{\pi}{4}$, we have:

Since $2\theta$ lies in $[0, \frac{\pi}{2}]$, which is a subset of $[-\frac{\pi}{2}, \frac{\pi}{2}]$, we have $\sin^{-1}(\sin(2\theta)) = 2\theta$.

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

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

This holds for $\frac{1}{\sqrt{2}} \leq x \leq 1$.

Let the given expression be $y = \tan^{-1} \left( \frac{\cos x}{1-\sin x} \right)$.

We will simplify the argument of the $\tan^{-1}$ function using half-angle identities.

Recall the identities: $\cos x = \cos^2\left(\frac{x}{2}\right) - \sin^2\left(\frac{x}{2}\right)$ and $1 = \cos^2\left(\frac{x}{2}\right) + \sin^2\left(\frac{x}{2}\right)$ and $\sin x = 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$.

The argument is:

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

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

The given range for $x$ is $-\frac{3\pi}{2} < x < \frac{\pi}{2}$.

Dividing by 2, we get $-\frac{3\pi}{4} < \frac{x}{2} < \frac{\pi}{4}$.

In the interval $(-\frac{3\pi}{4}, \frac{\pi}{4})$, $\tan\left(\frac{x}{2}\right) \neq 1$. This implies $\cos\left(\frac{x}{2}\right) - \sin\left(\frac{x}{2}\right) \neq 0$.

So, we can cancel the term $\left(\cos\left(\frac{x}{2}\right) - \sin\left(\frac{x}{2}\right)\right)$ from the numerator and the denominator:

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

Now, divide the numerator and the denominator by $\cos\left(\frac{x}{2}\right)$ (Note that $\cos\left(\frac{x}{2}\right) \neq 0$ for $-\frac{3\pi}{4} < \frac{x}{2} < \frac{\pi}{4}$ except possibly at $\frac{x}{2} = -\frac{\pi}{2}$, i.e., $x=-\pi$. At $x=-\pi$, $\frac{\cos(-\pi)}{1-\sin(-\pi)} = \frac{-1}{1-0} = \frac{-1}{1} = -1$. Our simplified form should also give $\tan\left(\frac{\pi}{4} - \frac{\pi}{2}\right) = \tan\left(-\frac{\pi}{4}\right)=-1$).

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

Using the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ and $\tan\left(\frac{\pi}{4}\right) = 1$:

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

So, the given expression is $y = \tan^{-1} \left( \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) \right)$.

The principal value branch of $\tan^{-1} \phi$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$. The identity $\tan^{-1}(\tan \phi) = \phi$ is valid if $\phi \in (-\frac{\pi}{2}, \frac{\pi}{2})$.

We need to check if the argument $\frac{\pi}{4} + \frac{x}{2}$ lies in this interval for the given range of $x$.

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

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

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

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

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

Let $\phi = \frac{\pi}{4} + \frac{x}{2}$. We see that $\phi$ lies in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, which is the principal value branch of $\tan^{-1}$.

Therefore, $\tan^{-1}(\tan \phi) = \phi$ is valid.

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

The simplest form is $\frac{\pi}{4} + \frac{x}{2}$.

Solution:

Let the given expression be $y = \cot^{-1} \left( \frac{1}{\sqrt{x^{2}-1}} \right)$.

We are given that $x > 1$.

Consider the substitution $x = \sec \theta$.

Since $x > 1$, the value of $\theta = \sec^{-1} x$ must be in the principal value branch of $\sec^{-1}$, which is $[0, \pi] - \{\frac{\pi}{2}\}$.

For $x > 1$, $\sec \theta > 1$, which means $\theta$ lies in the interval $(0, \frac{\pi}{2})$.

Substitute $x = \sec \theta$ into the expression inside the $\cot^{-1}$ function:

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

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

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

Since $\theta \in (0, \frac{\pi}{2})$, $\tan\theta$ is positive. So, $|\tan\theta| = \tan\theta$.

$= \frac{1}{\tan\theta} = \cot\theta$.

Now, substitute this back into the original expression for $y$:

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

The principal value branch of $\cot^{-1}\phi$ is $(0, \pi)$.

For the identity $\cot^{-1}(\cot\phi) = \phi$ to be valid, $\phi$ must lie in the interval $(0, \pi)$.

In our case, $\phi = \theta$. We found that $\theta \in (0, \frac{\pi}{2})$.

The interval $(0, \frac{\pi}{2})$ is a subset of $(0, \pi)$. Therefore, the identity is valid for our range of $\theta$.

$y = \theta$.

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

$y = \sec^{-1} x$.

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

Alternate Solution:

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

Consider the substitution $x = \text{cosec } \theta$.

Since $x > 1$, the value of $\theta = \text{cosec}^{-1} x$ must be in the principal value branch of $\text{cosec}^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$.

For $x > 1$, $\text{cosec } \theta > 1$, which means $\theta$ lies in the interval $(0, \frac{\pi}{2})$.

Substitute $x = \text{cosec } \theta$ into the expression inside the $\cot^{-1}$ function:

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

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

$= \frac{1}{\sqrt{\cot^{2}\theta}} = \frac{1}{|\cot\theta|}$.

Since $\theta \in (0, \frac{\pi}{2})$, $\cot\theta$ is positive. So, $|\cot\theta| = \cot\theta$.

$= \frac{1}{\cot\theta} = \tan\theta$.

Now, substitute this back into the original expression for $y$:

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

Use the identity $\tan\theta = \cot(\frac{\pi}{2} - \theta)$.

$y = \cot^{-1}\left(\cot\left(\frac{\pi}{2} - \theta\right)\right)$.

The principal value branch of $\cot^{-1}\phi$ is $(0, \pi)$.

For the identity $\cot^{-1}(\cot\phi) = \phi$ to be valid, $\phi$ must lie in the interval $(0, \pi)$.

In our case, $\phi = \frac{\pi}{2} - \theta$. We know that $\theta \in (0, \frac{\pi}{2})$.

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

Multiply by -1: $-\frac{\pi}{2} < -\theta < 0$.

Add $\frac{\pi}{2}$: $0 < \frac{\pi}{2} - \theta < \frac{\pi}{2}$.

The interval $(0, \frac{\pi}{2})$ is a subset of $(0, \pi)$. Therefore, the identity is valid for our range of $\phi$.

$y = \frac{\pi}{2} - \theta$.

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

$y = \frac{\pi}{2} - \text{cosec}^{-1} x$.

We know the identity $\sec^{-1} x + \text{cosec}^{-1} x = \frac{\pi}{2}$ for $|x| \geq 1$.

For $x > 1$, this identity holds.

Rearranging, we get $\sec^{-1} x = \frac{\pi}{2} - \text{cosec}^{-1} x$.

Thus, both simplified forms $\sec^{-1} x$ and $\frac{\pi}{2} - \text{cosec}^{-1} x$ are equivalent.

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

Let $x = \sin \theta$.

Since $x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$, and the principal value branch of $\sin^{-1}x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, $\theta = \sin^{-1} x$ must lie in this range.

For $x = -\frac{1}{2}$, $\sin \theta = -\frac{1}{2}$, which gives $\theta = -\frac{\pi}{6}$ in the principal branch.

For $x = \frac{1}{2}$, $\sin \theta = \frac{1}{2}$, which gives $\theta = \frac{\pi}{6}$ in the principal branch.

Thus, for $x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$, we have $\theta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right]$.

Consider the Right Hand Side (RHS) of the identity: $\sin^{-1} (3x – 4x^{3})$.

Substitute $x = \sin \theta$ into the expression $3x - 4x^3$:

$3x - 4x^3 = 3\sin \theta - 4\sin^3 \theta$.

Using the trigonometric identity $\sin(3\theta) = 3\sin \theta - 4\sin^3 \theta$, we have:

$3x - 4x^3 = \sin(3\theta)$.

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

For the identity $\sin^{-1}(\sin \phi) = \phi$ to hold, the angle $\phi$ must be in the principal value branch of $\sin^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

In this case, $\phi = 3\theta$. We need to check if $3\theta$ lies in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ based on the range of $\theta$ in (i).

From (i), $-\frac{\pi}{6} \leq \theta \leq \frac{\pi}{6}$.

Multiplying the inequality by 3:

From (ii), the angle $3\theta$ lies in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$, which is the principal value branch of $\sin^{-1}$.

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

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

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

This is equal to the Left Hand Side (LHS).

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

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

Let $x = \cos \theta$.

Since $x \in \left[\frac{1}{2}, 1\right]$, and the principal value branch of $\cos^{-1}x$ is $[0, \pi]$, $\theta = \cos^{-1} x$ must lie in this range.

For $x = 1$, $\cos \theta = 1$, which gives $\theta = 0$ in the principal branch.

For $x = \frac{1}{2}$, $\cos \theta = \frac{1}{2}$, which gives $\theta = \frac{\pi}{3}$ in the principal branch.

Since $\cos \theta$ is a decreasing function on $[0, \pi]$, for $x \in \left[\frac{1}{2}, 1\right]$, we have $\theta = \cos^{-1} x \in \left[0, \frac{\pi}{3}\right]$.

Consider the Right Hand Side (RHS) of the identity: $\cos^{-1} (4x^{3} – 3x)$.

Substitute $x = \cos \theta$ into the expression $4x^3 - 3x$:

$4x^3 - 3x = 4\cos^3 \theta - 3\cos \theta$.

Using the trigonometric identity $\cos(3\theta) = 4\cos^3 \theta - 3\cos \theta$, we have:

$4x^3 - 3x = \cos(3\theta)$.

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

For the identity $\cos^{-1}(\cos \phi) = \phi$ to hold, the angle $\phi$ must be in the principal value branch of $\cos^{-1}$, which is $[0, \pi]$.

In this case, $\phi = 3\theta$. We need to check if $3\theta$ lies in $[0, \pi]$ based on the range of $\theta$ in (i).

From (i), $0 \leq \theta \leq \frac{\pi}{3}$.

Multiplying the inequality by 3:

From (ii), the angle $3\theta$ lies in the interval $[0, \pi]$, which is the principal value branch of $\cos^{-1}$.

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

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

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

This is equal to the Left Hand Side (LHS).

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

Let the given expression be $y = \tan^{-1} \left( \frac{\sqrt{1+x^{2}}-1}{x} \right)$.

We are given that $x \neq 0$.

Consider the substitution $x = \tan \theta$.

Since $x \neq 0$, $\tan \theta \neq 0$, which implies $\theta \neq n\pi$ for any integer $n$.

If we consider $\theta = \tan^{-1} x$, the principal value branch of $\tan^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

So, $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$ and $\theta \neq 0$.

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 $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$, $\cos \theta > 0$, so $\sec \theta = \frac{1}{\cos \theta} > 0$. Thus, $|\sec\theta| = \sec\theta$.

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

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

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

Use the half-angle identities: $1-\cos \theta = 2\sin^{2}\left(\frac{\theta}{2}\right)$ and $\sin \theta = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)$.

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

Since $x \neq 0$, $\theta \neq 0$. Also $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$, so $\frac{\theta}{2} \in (-\frac{\pi}{4}, \frac{\pi}{4})$ and $\frac{\theta}{2} \neq 0$. Thus, $\sin\left(\frac{\theta}{2}\right) \neq 0$. We can cancel $\sin\left(\frac{\theta}{2}\right)$.

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

So, the given expression is $y = \tan^{-1} \left( \tan\left(\frac{\theta}{2}\right) \right)$.

The principal value branch of $\tan^{-1}\phi$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$. The identity $\tan^{-1}(\tan \phi) = \phi$ is valid if $\phi \in (-\frac{\pi}{2}, \frac{\pi}{2})$.

In our case, $\phi = \frac{\theta}{2}$. We know that $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$ and $\theta \neq 0$.

Multiplying the inequality $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ by $\frac{1}{2}$, we get $-\frac{\pi}{4} < \frac{\theta}{2} < \frac{\pi}{4}$.

The interval $(-\frac{\pi}{4}, \frac{\pi}{4})$ is a subset of $(-\frac{\pi}{2}, \frac{\pi}{2})$. Also, $\frac{\theta}{2} \neq 0$ since $\theta \neq 0$.

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

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

$y = \frac{1}{2}\tan^{-1} x$.

The simplest form is $\frac{1}{2}\tan^{-1} x$.

Let the given expression be $y = \tan^{-1} \left( \sqrt{\frac{1-\cos x}{1+\cos x}} \right)$.

We are given the condition $0 < x < \pi$.

We simplify the argument of the $\tan^{-1}$ function using half-angle identities.

Recall the identities: $1 - \cos x = 2\sin^2\left(\frac{x}{2}\right)$ and $1 + \cos x = 2\cos^2\left(\frac{x}{2}\right)$.

Substitute these into the expression under the square root:

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

Now, take the square root:

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

Consider the given domain for $x$: $0 < x < \pi$.

Dividing the inequality by 2, we get the range for $\frac{x}{2}$:

In the interval $(0, \frac{\pi}{2})$, the value of $\tan\left(\frac{x}{2}\right)$ is positive.

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

Substitute this back into the expression for $y$:

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

The principal value branch of $\tan^{-1}\phi$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$. The identity $\tan^{-1}(\tan \phi) = \phi$ holds true if $\phi$ lies within this interval.

In our case, $\phi = \frac{x}{2}$. From the range $0 < \frac{x}{2} < \frac{\pi}{2}$, we see that $\frac{x}{2}$ lies within the principal value branch $(-\frac{\pi}{2}, \frac{\pi}{2})$.

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

The simplest form of the given function is $\frac{x}{2}$.

Let the given expression be $y = \tan^{-1} \left( \frac{\cos x - \sin x}{\cos x + \sin x} \right)$.

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

We simplify the argument of the $\tan^{-1}$ function by dividing the numerator and the denominator by $\cos x$. Note that for the given range $-\frac{\pi}{4} < x < \frac{3\pi}{4}$, $\cos x = 0$ only at $x = \frac{\pi}{2}$. If $x = \frac{\pi}{2}$, the expression is $\tan^{-1}(\frac{0-1}{0+1}) = \tan^{-1}(-1) = -\frac{\pi}{4}$. Our final simplified form should yield the same value at $x=\frac{\pi}{2}$.

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

We use the identity $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.

Let $A = \frac{\pi}{4}$. Then $\tan A = \tan\left(\frac{\pi}{4}\right) = 1$. Let $B = x$.

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

Now, substitute this back into the expression for $y$:

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

The principal value branch of $\tan^{-1}\phi$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$. The identity $\tan^{-1}(\tan \phi) = \phi$ is valid if $\phi$ lies within this interval.

In our case, $\phi = \frac{\pi}{4} - x$. We need to check if this argument lies in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$ for the given range of $x$.

Given: $-\frac{\pi}{4} < x < \frac{3\pi}{4}$.

Multiply the inequality by -1 and reverse the inequality signs:

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

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

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

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

Therefore, the identity $\tan^{-1}(\tan \phi) = \phi$ is valid for $\phi = \frac{\pi}{4} - x$ in this range.

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

The simplest form of the given function is $\frac{\pi}{4} - x$.

Let the given expression be $y = \tan^{-1} \left( \frac{x}{\sqrt{a^{2}-x^{2}}} \right)$.

We are given the condition $|x| < a$. This implies $a > 0$ for the condition to be meaningful and for the expression $\sqrt{a^2-x^2}$ to be defined (since $a^2-x^2 > 0$).

Consider the substitution $x = a \sin \theta$.

Since $|x| < a$ and $a > 0$, we have $\frac{|x|}{a} < 1$, which means $\left|\frac{x}{a}\right| < 1$.

Let $\theta = \sin^{-1}\left(\frac{x}{a}\right)$. The principal value branch of $\sin^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. For $\frac{x}{a} \in (-1, 1)$, the value of $\theta = \sin^{-1}\left(\frac{x}{a}\right)$ lies in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Substitute $x = a \sin \theta$ into the expression inside the $\tan^{-1}$ function:

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

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

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

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

$= \frac{a \sin \theta}{|a \cos \theta|}$

Since $a > 0$, $|a| = a$.

$= \frac{a \sin \theta}{a |\cos \theta|}$

$= \frac{\sin \theta}{|\cos \theta|}$

From the range of $\theta$, $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$, we know that $\cos \theta > 0$.

So, $|\cos \theta| = \cos \theta$.

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

Now, substitute this back into the expression for $y$:

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

The principal value branch of $\tan^{-1}\phi$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$. The identity $\tan^{-1}(\tan \phi) = \phi$ is valid if $\phi$ lies within this interval.

In our case, $\phi = \theta$. We found that $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$.

This interval is exactly the principal value branch of $\tan^{-1}$.

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

Substitute back $\theta = \sin^{-1}\left(\frac{x}{a}\right)$:

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

The simplest form of the given function is $\sin^{-1}\left(\frac{x}{a}\right)$.

Let the given expression be $y = \tan^{-1} \left( \frac{3a^{2}x-x^{3}}{a^{3}-3ax^{2}} \right)$.

We are given the conditions $a > 0$ and $-\frac{a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$.

Consider the substitution $x = a \tan \theta$.

From the condition $-\frac{a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$, substitute $x = a \tan \theta$:

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

Since $a > 0$, we can divide by $a$ without changing the inequalities:

The principal value branch of $\tan^{-1}u$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$, and $\tan \theta$ is increasing on this interval.

Since $\tan\left(-\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}$ and $\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$, the inequality $-\frac{1}{\sqrt{3}} < \tan \theta < \frac{1}{\sqrt{3}}$ implies:

From $x = a \tan \theta$, we have $\frac{x}{a} = \tan \theta$, so $\theta = \tan^{-1}\left(\frac{x}{a}\right)$. Note that this is valid because of the range of $\theta$ in (i) which is within the principal branch $(-\frac{\pi}{2}, \frac{\pi}{2})$.

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

$\frac{3a^{2}x-x^{3}}{a^{3}-3ax^{2}} = \frac{3a^{2}(a \tan \theta)-(a \tan \theta)^{3}}{a^{3}-3a(a \tan \theta)^{2}}$

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

$= \frac{3a^{3} \tan \theta - a^{3} \tan^{3}\theta}{a^{3}-3a^{3} \tan^{2}\theta}$

Factor out $a^3$ from the numerator and the denominator:

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

Since $a > 0$, $a^3 \neq 0$, so we can cancel $a^3$:

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

Using the trigonometric identity $\tan(3\phi) = \frac{3\tan\phi - \tan^3\phi}{1 - 3\tan^2\phi}$, with $\phi = \theta$, we have:

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

Now, substitute this back into the original expression for $y$:

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

The identity $\tan^{-1}(\tan \phi) = \phi$ holds true if $\phi$ lies within the principal value branch of $\tan^{-1}$, which is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

In our case, $\phi = 3\theta$. We check the range of $3\theta$ from (i):

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

Multiply by 3:

The range of $3\theta$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$, which is the principal value branch of $\tan^{-1}$.

Therefore, the identity $\tan^{-1}(\tan \phi) = \phi$ is valid for $\phi = 3\theta$ in this range.

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

Substitute back $\theta = \tan^{-1}\left(\frac{x}{a}\right)$:

$y = 3\tan^{-1}\left(\frac{x}{a}\right)$.

The simplest form of the given function is $3\tan^{-1}\left(\frac{x}{a}\right)$.

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

We evaluate the expression by working from the innermost function outwards.

First, consider the innermost term $\sin^{-1} \frac{1}{2}$.

Let $y = \sin^{-1} \frac{1}{2}$. This means $\sin y = \frac{1}{2}$.

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

The angle in this interval whose sine is $\frac{1}{2}$ is $\frac{\pi}{6}$.

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

Next, evaluate $2 \sin^{-1} \frac{1}{2}$.

$2 \times \sin^{-1} \frac{1}{2} = 2 \times \frac{\pi}{6} = \frac{\pi}{3}$.

Now, evaluate $\cos\left( 2\sin^{-1} \frac{1}{2}\right)$, which is $\cos\left(\frac{\pi}{3}\right)$.

Next, evaluate $2 \cos\left( 2\sin^{-1} \frac{1}{2}\right)$, which is $2 \times \frac{1}{2}$.

Finally, evaluate the outermost function $\tan^{-1} \left[2 \cos\left( 2\sin^{-1} \frac{1}{2}\right) \right]$, which is $\tan^{-1}(1)$.

Let $z = \tan^{-1}(1)$. This means $\tan z = 1$.

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

The angle in this interval whose tangent is 1 is $\frac{\pi}{4}$.

Since $\frac{\pi}{4} \in (-\frac{\pi}{2}, \frac{\pi}{2})$.

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

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

We are given the conditions $|x| < 1$, $y > 0$, and $xy < 1$.

We use the following standard identities:

For $|x| \leq 1$, $\sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x$. The given condition $|x| < 1$ satisfies this range.

For $y \geq 0$, $\cos^{-1}\left(\frac{1-y^2}{1+y^2}\right) = 2\tan^{-1}y$. The given condition $y > 0$ satisfies this range.

Substitute these identities into the given expression:

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

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

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

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

In our case, $A=x$ and $B=y$. The given condition is $xy < 1$, so the identity is applicable.

Substitute this back into the expression:

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

Using the property $\tan(\tan^{-1} Z) = Z$ for all real numbers $Z$. Here $Z = \frac{x+y}{1-xy}$. Since $xy < 1$, $1-xy \neq 0$, so $Z$ is a real number.

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

The value of the expression is $\frac{x+y}{1-xy}$.

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

We know the property $\sin^{-1}(\sin \theta) = \theta$ if $\theta$ lies in the principal value branch of $\sin^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

The angle given is $\frac{2\pi}{3}$. Let's check if this angle is within the principal value branch:

Since $120^\circ$ is not in the interval $[-90^\circ, 90^\circ]$, the property $\sin^{-1}(\sin \theta) = \theta$ cannot be applied directly.

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

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

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

Now, the expression becomes $\sin^{-1}\left(\sin\left(\frac{\pi}{3}\right)\right)$.

The angle is now $\frac{\pi}{3}$. Let's check if this angle is in the principal value branch of $\sin^{-1}$:

Since $60^\circ$ is in the interval $[-90^\circ, 90^\circ]$, the property $\sin^{-1}(\sin \phi) = \phi$ can be applied.

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

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

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

We know the property $\tan^{-1}(\tan \theta) = \theta$ if $\theta$ lies in the principal value branch of $\tan^{-1}$, which is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

The angle given is $\frac{3\pi}{4}$. Let's check if this angle is within the principal value branch:

Since $135^\circ$ is not in the interval $(-90^\circ, 90^\circ)$, the property $\tan^{-1}(\tan \theta) = \theta$ cannot be applied directly.

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

We know that $\tan \frac{3\pi}{4} = \tan\left(\pi - \frac{\pi}{4}\right)$.

Using the identity $\tan(\pi - \theta) = -\tan \theta$:

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

Now we need to find an angle $\phi \in (-\frac{\pi}{2}, \frac{\pi}{2})$ such that $\tan \phi = -1$.

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

So, the expression becomes $\tan^{-1}\left(\tan\left(-\frac{\pi}{4}\right)\right)$.

The angle is now $-\frac{\pi}{4}$. Let's check if this angle is in the principal value branch of $\tan^{-1}$:

Since $-45^\circ$ is in the interval $(-90^\circ, 90^\circ)$, the property $\tan^{-1}(\tan \phi) = \phi$ can be applied.

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

The value of the expression is $-\frac{\pi}{4}$.

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

We can rewrite the expression in the form $\tan(A+B)$, where $A = \sin^{-1} \frac{3}{5}$ and $B = \cot^{-1}\frac{3}{2}$.

To evaluate $\tan(A+B)$, we need to find $\tan A$ and $\tan B$.

Consider $A = \sin^{-1} \frac{3}{5}$. Let $\theta = \sin^{-1} \frac{3}{5}$.

Then $\sin \theta = \frac{3}{5}$. Since $\frac{3}{5} > 0$, the principal value $\theta$ lies in $(0, \frac{\pi}{2})$.

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

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

Since $\theta \in (0, \frac{\pi}{2})$, $\cos \theta$ is positive. So, $\cos \theta = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

Now, $\tan A = \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3/5}{4/5} = \frac{3}{4}$.

Consider $B = \cot^{-1}\frac{3}{2}$. Let $\phi = \cot^{-1}\frac{3}{2}$.

Then $\cot \phi = \frac{3}{2}$. Since $\frac{3}{2} > 0$, the principal value $\phi$ lies in $(0, \frac{\pi}{2})$.

We can find $\tan \phi$ using the identity $\tan \phi = \frac{1}{\cot \phi}$.

$\tan B = \tan \phi = \frac{1}{3/2} = \frac{2}{3}$.

The expression is $\tan(A+B)$. We use the trigonometric identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.

Substitute the values of $\tan A$ and $\tan B$:

$\tan(A+B) = \frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \times \frac{2}{3}}$

Calculate the numerator: $\frac{3}{4} + \frac{2}{3} = \frac{3 \times 3 + 4 \times 2}{12} = \frac{9 + 8}{12} = \frac{17}{12}$.

Calculate the denominator: $1 - \frac{\cancel{3}\times\cancel{2}}{\cancel{4}_{2}\times\cancel{3}} = 1 - \frac{1}{2} = \frac{2-1}{2} = \frac{1}{2}$.

So, $\tan(A+B) = \frac{\frac{17}{12}}{\frac{1}{2}}$.

$= \frac{17}{12} \times \frac{2}{1} = \frac{17 \times \cancel{2}^{1}}{\cancel{12}_{6}} = \frac{17}{6}$.

Alternatively, we can convert both inverse functions to $\tan^{-1}$ first.

From the calculations above, $A = \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{3}{4}$.

And $B = \cot^{-1} \frac{3}{2} = \tan^{-1} \frac{2}{3}$.

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

Use the identity $\tan^{-1} X + \tan^{-1} Y = \tan^{-1}\left(\frac{X+Y}{1-XY}\right)$ for $XY < 1$.

Here $X = \frac{3}{4}$ and $Y = \frac{2}{3}$. Their product $XY = \frac{3}{4} \times \frac{2}{3} = \frac{1}{2}$, which is less than 1.

So, $\tan^{-1} \frac{3}{4} + \tan^{-1} \frac{2}{3} = \tan^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1 - \frac{3}{4}\times\frac{2}{3}}\right) = \tan^{-1}\left(\frac{\frac{17}{12}}{\frac{1}{2}}\right) = \tan^{-1}\left(\frac{17}{6}\right)$.

The expression becomes $\tan\left(\tan^{-1}\left(\frac{17}{6}\right)\right)$.

Using the property $\tan(\tan^{-1} Z) = Z$ for any real number $Z$, we get:

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

The value of the expression is $\frac{17}{6}$.

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

We know the property $\cos^{-1}(\cos \theta) = \theta$ if $\theta$ lies in the principal value branch of $\cos^{-1}$, which is $[0, \pi]$.

The angle given is $\frac{7\pi}{6}$. Let's check if this angle is within the principal value branch:

Since $210^\circ$ is not in the interval $[0^\circ, 180^\circ]$, the property $\cos^{-1}(\cos \theta) = \theta$ cannot be applied directly.

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

We know that the cosine function is positive in the first and fourth quadrants and negative in the second and third quadrants.

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

The reference angle is $\frac{7\pi}{6} - \pi = \frac{\pi}{6}$. So, $\cos \frac{7\pi}{6} = -\cos \frac{\pi}{6}$.

We need an angle $\phi \in [0, \pi]$ such that $\cos \phi = -\cos \frac{\pi}{6}$.

In the second quadrant, $\cos(\pi - \theta) = -\cos \theta$.

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

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

So, $\cos \frac{7\pi}{6} = \cos \frac{5\pi}{6}$.

Now, the expression becomes $\cos^{-1}\left(\cos\left(\frac{5\pi}{6}\right)\right)$.

The angle is now $\frac{5\pi}{6}$. Let's check if this angle is in the principal value branch of $\cos^{-1}$:

Since $150^\circ$ is in the interval $[0^\circ, 180^\circ]$, the property $\cos^{-1}(\cos \phi) = \phi$ can be applied.

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

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

We evaluate the expression by working from the innermost function outwards.

First, consider the innermost term $\sin^{-1} \left( -\frac{1}{2} \right)$.

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

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

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

Since $\sin y$ is negative, $y$ must be in the interval $[-\frac{\pi}{2}, 0]$.

Using the identity $\sin(-\theta) = -\sin\theta$, we have $\sin y = -\sin\left(\frac{\pi}{6}\right) = \sin\left(-\frac{\pi}{6}\right)$.

The angle $-\frac{\pi}{6}$ lies in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Next, substitute this value into the argument of the outer sine function: $\frac{\pi}{3}-\sin^{-1} \left( -\frac{1}{2} \right)$.

Argument $= \frac{\pi}{3} - \left(-\frac{\pi}{6}\right) = \frac{\pi}{3} + \frac{\pi}{6}$.

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

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

Finally, evaluate the sine of this angle: $\sin\left(\frac{\pi}{2}\right)$.

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

We find the principal value of each inverse trigonometric function separately.

Consider the first term, $\tan^{-1} \sqrt{3}$.

Let $y_1 = \tan^{-1} \sqrt{3}$. This means $\tan y_1 = \sqrt{3}$.

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

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

The angle $\frac{\pi}{3}$ lies in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Consider the second term, $\cot^{-1} (-\sqrt{3})$.

Let $y_2 = \cot^{-1} (-\sqrt{3})$. This means $\cot y_2 = -\sqrt{3}$.

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

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

Since $\cot y_2$ is negative, $y_2$ must lie in the second quadrant within the principal value branch $(0, \pi)$.

Using the identity $\cot(\pi - \theta) = -\cot \theta$:

The angle $\frac{5\pi}{6}$ lies in the interval $(0, \pi)$.

Now, we substitute the principal values from (i) and (ii) into the given expression:

$\tan^{-1} \sqrt{3} - \cot^{-1} (-\sqrt{3}) = \frac{\pi}{3} - \frac{5\pi}{6}$.

To subtract these fractions, find a common denominator, which is 6.

$= \frac{2\pi}{6} - \frac{5\pi}{6}$

$= \frac{2\pi - 5\pi}{6}$

$= \frac{-3\pi}{6}$

Simplify the fraction:

$= -\frac{\cancel{3}^{1}\pi}{\cancel{6}_{2}} = -\frac{\pi}{2}$.

The value of the expression is $-\frac{\pi}{2}$.

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

Solution:

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

We know the property $\sin^{-1}(\sin \theta) = \theta$ if and only if $\theta$ lies in the principal value branch of $\sin^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

The given angle is $\theta = \frac{3\pi}{5}$.

Let's check if $\frac{3\pi}{5}$ is in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$:

Since $108^\circ$ is not in the interval $[-90^\circ, 90^\circ]$, we cannot directly say $\sin^{-1}\left(\sin\frac{3\pi}{5}\right) = \frac{3\pi}{5}$.

We need to find an angle $\phi$ in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $\sin \phi = \sin \frac{3\pi}{5}$.

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

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

Calculate the angle $\pi - \frac{3\pi}{5}$:

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

So, we have $\sin \frac{3\pi}{5} = \sin \frac{2\pi}{5}$.

Now the expression is $\sin^{-1}\left(\sin\left(\frac{2\pi}{5}\right)\right)$.

The new angle is $\phi = \frac{2\pi}{5}$. Let's check if this angle is in the principal value branch of $\sin^{-1}$:

Since $72^\circ$ is in the interval $[-90^\circ, 90^\circ]$, the property $\sin^{-1}(\sin \phi) = \phi$ is valid for $\phi = \frac{2\pi}{5}$.

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

The value of the expression is $\frac{2\pi}{5}$.

We need to find the value of $\cos^{-1} \left(\cos \frac{13\pi}{6}\right)$.

We know the property $\cos^{-1}(\cos \theta) = \theta$ if $\theta$ lies in the principal value branch of $\cos^{-1}$, which is $[0, \pi]$.

The given angle is $\theta = \frac{13\pi}{6}$.

Let's check if $\frac{13\pi}{6}$ is in the interval $[0, \pi]$:

Since $390^\circ$ is not in the interval $[0^\circ, 180^\circ]$, we cannot directly say $\cos^{-1}\left(\cos\frac{13\pi}{6}\right) = \frac{13\pi}{6}$.

We need to find an angle $\phi$ in the interval $[0, \pi]$ such that $\cos \phi = \cos \frac{13\pi}{6}$.

The cosine function has a period of $2\pi$. We can write the given angle as $\frac{13\pi}{6} = \frac{12\pi + \pi}{6} = 2\pi + \frac{\pi}{6}$.

Using the identity $\cos(2n\pi + \theta) = \cos \theta$ for any integer $n$, we have:

So, the expression becomes $\cos^{-1}\left(\cos\left(\frac{\pi}{6}\right)\right)$.

The new angle is $\phi = \frac{\pi}{6}$. Let's check if this angle is in the principal value branch of $\cos^{-1}$:

Since $30^\circ$ is in the interval $[0^\circ, 180^\circ]$, the property $\cos^{-1}(\cos \phi) = \phi$ is valid for $\phi = \frac{\pi}{6}$.

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

The value of the expression is $\frac{\pi}{6}$.

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

We know the property $\tan^{-1}(\tan \theta) = \theta$ if $\theta$ lies in the principal value branch of $\tan^{-1}$, which is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

The given angle is $\theta = \frac{7\pi}{6}$.

Let's check if $\frac{7\pi}{6}$ is in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$:

Since $210^\circ$ is not in the interval $(-90^\circ, 90^\circ)$, we cannot directly say $\tan^{-1}\left(\tan\frac{7\pi}{6}\right) = \frac{7\pi}{6}$.

We need to find an angle $\phi$ in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$ such that $\tan \phi = \tan \frac{7\pi}{6}$.

The tangent function has a period of $\pi$. We can write the given angle as $\frac{7\pi}{6} = \pi + \frac{\pi}{6}$.

Using the identity $\tan(\pi + \theta) = \tan \theta$, we have:

So, the expression becomes $\tan^{-1}\left(\tan\left(\frac{\pi}{6}\right)\right)$.

The new angle is $\phi = \frac{\pi}{6}$. Let's check if this angle is in the principal value branch of $\tan^{-1}$:

Since $30^\circ$ is in the interval $(-90^\circ, 90^\circ)$, the property $\tan^{-1}(\tan \phi) = \phi$ is valid for $\phi = \frac{\pi}{6}$.

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

The value of the expression is $\frac{\pi}{6}$.

To Prove: $2\sin^{-1} \frac{3}{5} = \tan^{-1} \frac{24}{7}$.

Let the Left Hand Side (LHS) be $2\sin^{-1} \frac{3}{5}$.

Let $\theta = \sin^{-1} \frac{3}{5}$.

By definition, $\sin \theta = \frac{3}{5}$.

Since the argument $\frac{3}{5}$ is positive and $0 < \frac{3}{5} \leq 1$, the principal value of $\theta = \sin^{-1} \frac{3}{5}$ lies in the interval $(0, \frac{\pi}{2}]$.

Also, since $\frac{3}{5} < \frac{1}{\sqrt{2}}$ ($\frac{3}{5} = 0.6$, $\frac{1}{\sqrt{2}} \approx 0.707$), and $\sin x$ is increasing on $[0, \frac{\pi}{2}]$, we have $\sin^{-1} \frac{3}{5} < \sin^{-1} \frac{1}{\sqrt{2}} = \frac{\pi}{4}$.

Now, consider the angle $2\theta$. Multiplying the inequality by 2:

We need to show that $2\theta$ is equal to $\tan^{-1} \frac{24}{7}$. Both angles are in the interval $(0, \frac{\pi}{2})$. If two angles in this interval have the same tangent, they are equal.

Let's find $\tan(2\theta)$. We use the double angle formula for tangent:

We need to find $\tan \theta$. We know $\sin \theta = \frac{3}{5}$ and $\theta \in (0, \frac{\pi}{4})$.

Using the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$, we first find $\cos \theta$. Since $\theta \in (0, \frac{\pi}{4})$, $\cos \theta > 0$.

Now find $\tan \theta$:

Substitute this value into the expression for $\tan(2\theta)$:

So, we have $\tan(2\theta) = \frac{24}{7}$.

Since $0 < 2\theta < \frac{\pi}{2}$, we can apply the $\tan^{-1}$ function to both sides:

As $2\theta$ is within the principal value branch of $\tan^{-1}$, $\tan^{-1}(\tan(2\theta)) = 2\theta$.

Substitute back $\theta = \sin^{-1} \frac{3}{5}$:

This is the Right Hand Side (RHS) of the identity.

Thus, $2\sin^{-1} \frac{3}{5} = \tan^{-1} \frac{24}{7}$ is proved.

To Prove: $\sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{77}{36}$.

Consider the Left Hand Side (LHS): $\sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5}$.

We will convert each $\sin^{-1}$ term into a $\tan^{-1}$ term.

Let $\alpha = \sin^{-1} \frac{8}{17}$. By definition, $\sin \alpha = \frac{8}{17}$.

Since $0 < \frac{8}{17} < 1$, the principal value of $\alpha$ lies in the interval $(0, \frac{\pi}{2})$.

For $\alpha \in (0, \frac{\pi}{2})$, $\cos \alpha > 0$. We can find $\cos \alpha$ using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$:

Now, we find $\tan \alpha$:

Since $\alpha \in (0, \frac{\pi}{2})$, $\tan \alpha$ is positive. The angle $\alpha$ is also in the principal value branch of $\tan^{-1}$.

Let $\beta = \sin^{-1} \frac{3}{5}$. By definition, $\sin \beta = \frac{3}{5}$.

Since $0 < \frac{3}{5} < 1$, the principal value of $\beta$ lies in the interval $(0, \frac{\pi}{2})$.

For $\beta \in (0, \frac{\pi}{2})$, $\cos \beta > 0$. We can find $\cos \beta$ using the identity $\sin^2 \beta + \cos^2 \beta = 1$:

Now, we find $\tan \beta$:

Since $\beta \in (0, \frac{\pi}{2})$, $\tan \beta$ is positive. The angle $\beta$ is also in the principal value branch of $\tan^{-1}$.

Now, substitute these back into the LHS:

LHS $= \alpha + \beta = \tan^{-1} \frac{8}{15} + \tan^{-1} \frac{3}{4}$.

We use the identity $\tan^{-1} X + \tan^{-1} Y = \tan^{-1}\left(\frac{X+Y}{1-XY}\right)$, which is valid for $X > 0$, $Y > 0$, and $XY < 1$.

Here, $X = \frac{8}{15}$ and $Y = \frac{3}{4}$. Both are positive.

Check the product $XY$: $XY = \frac{8}{15} \times \frac{3}{4} = \frac{24}{60} = \frac{2}{5}$.

Since $XY = \frac{2}{5} < 1$, the identity is applicable.

Calculate $X+Y$ and $1-XY$:

Now, calculate $\frac{X+Y}{1-XY}$:

So, LHS $= \tan^{-1}\left(\frac{77}{36}\right)$.

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

Thus, $\sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{77}{36}$ is proved.

To Prove: $\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1} \frac{33}{65}$.

Consider the Left Hand Side (LHS): $\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13}$.

Let $x = \frac{4}{5}$ and $y = \frac{12}{13}$.

We use the identity $\cos^{-1} x + \cos^{-1} y = \cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2})$.

This identity is valid for $x, y \in [0, 1]$ and $x+y \geq 0$. If $x, y \geq 0$, the condition $x+y \geq 0$ is always satisfied.

Check the conditions for $x = \frac{4}{5}$ and $y = \frac{12}{13}$:

Let $A = \cos^{-1} \frac{4}{5}$ and $B = \cos^{-1} \frac{12}{13}$. Since $0 < \frac{4}{5} < 1$ and $0 < \frac{12}{13} < 1$, we have $A \in (0, \frac{\pi}{2})$ and $B \in (0, \frac{\pi}{2})$. Thus, $A+B \in (0, \pi)$, which is within the principal value branch of $\cos^{-1}$. So the identity form $\cos^{-1}(Z)$ will correctly give $A+B$.

Calculate $\sqrt{1-x^2}$ and $\sqrt{1-y^2}$:

Now, calculate the argument of the resulting $\cos^{-1}$ using the identity:

So, applying the identity:

This is the Right Hand Side (RHS).

Thus, $\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1} \frac{33}{65}$ is proved.

To Prove: $\cos^{-1} \frac{12}{13} + \sin^{-1} \frac{3}{5} = \sin^{-1} \frac{56}{65}$.

Consider the Left Hand Side (LHS): $\cos^{-1} \frac{12}{13} + \sin^{-1} \frac{3}{5}$.

Let $A = \cos^{-1} \frac{12}{13}$. By definition, $\cos A = \frac{12}{13}$.

Since $0 < \frac{12}{13} < 1$, the principal value of $A = \cos^{-1} \frac{12}{13}$ lies in the interval $(0, \frac{\pi}{2})$.

For $A \in (0, \frac{\pi}{2})$, $\sin A > 0$. We find $\sin A$ using the identity $\sin^2 A + \cos^2 A = 1$:

Let $B = \sin^{-1} \frac{3}{5}$. By definition, $\sin B = \frac{3}{5}$.

Since $0 < \frac{3}{5} < 1$, the principal value of $B = \sin^{-1} \frac{3}{5}$ lies in the interval $(0, \frac{\pi}{2})$.

For $B \in (0, \frac{\pi}{2})$, $\cos B > 0$. We find $\cos B$ using the identity $\sin^2 B + \cos^2 B = 1$:

The LHS is $A+B$. We want to show that $A+B = \sin^{-1} \frac{56}{65}$.

Let's evaluate $\sin(A+B)$ using the sum formula for sine:

Substitute the values of $\sin A, \cos B, \cos A, \sin B$:

Now we need to determine if $A+B$ is equal to the principal value $\sin^{-1} \frac{56}{65}$. This requires checking if $A+B$ lies in the principal value branch of $\sin^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

From the ranges of A and B:

Adding these inequalities:

We need to show that $A+B \leq \frac{\pi}{2}$. This is equivalent to showing $A \leq \frac{\pi}{2} - B$.

Since $A \in (0, \frac{\pi}{2})$ and $B \in (0, \frac{\pi}{2})$, $\frac{\pi}{2}-B \in (0, \frac{\pi}{2})$.

In the interval $(0, \frac{\pi}{2})$, the cosine function is strictly decreasing. So, $A \leq \frac{\pi}{2} - B$ is equivalent to $\cos A \geq \cos(\frac{\pi}{2} - B) = \sin B$.

We check if $\cos A \geq \sin B$:

To compare these fractions, we can cross-multiply: $12 \times 5$ vs $3 \times 13$.

Since $60 \geq 39$ is true, $\frac{12}{13} \geq \frac{3}{5}$ is true. Thus, $\cos A \geq \sin B$, which implies $A \leq \frac{\pi}{2} - B$, or $A+B \leq \frac{\pi}{2}$.

So we have $0 < A+B \leq \frac{\pi}{2}$. This interval is a subset of the principal value branch of $\sin^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Since $\sin(A+B) = \frac{56}{65}$ and $A+B$ is in the principal value branch of $\sin^{-1}$, we can take the inverse sine of both sides:

Substitute back $A = \cos^{-1} \frac{12}{13}$ and $B = \sin^{-1} \frac{3}{5}$:

This is the Right Hand Side (RHS).

Thus, $\cos^{-1} \frac{12}{13} + \sin^{-1} \frac{3}{5} = \sin^{-1} \frac{56}{65}$ is proved.

To Prove: $\tan^{-1} \frac{63}{16} = \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}$.

Consider the Right Hand Side (RHS): $\sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}$.

We will convert each term into a $\tan^{-1}$ term.

Let $A = \sin^{-1} \frac{5}{13}$.

By definition, $\sin A = \frac{5}{13}$.

Since $0 < \frac{5}{13} < 1$, the principal value of $A = \sin^{-1} \frac{5}{13}$ lies in the interval $(0, \frac{\pi}{2})$.

For $A \in (0, \frac{\pi}{2})$, $\cos A > 0$. We find $\cos A$ using the identity $\sin^2 A + \cos^2 A = 1$:

Now, we find $\tan A$:

Since $A \in (0, \frac{\pi}{2})$, $\tan A$ is positive. The angle $A$ is in the principal value branch of $\tan^{-1}$.

Let $B = \cos^{-1} \frac{3}{5}$.

By definition, $\cos B = \frac{3}{5}$.

Since $0 < \frac{3}{5} < 1$, the principal value of $B = \cos^{-1} \frac{3}{5}$ lies in the interval $(0, \frac{\pi}{2})$.

For $B \in (0, \frac{\pi}{2})$, $\sin B > 0$. We find $\sin B$ using the identity $\sin^2 B + \cos^2 B = 1$:

Now, we find $\tan B$:

Since $B \in (0, \frac{\pi}{2})$, $\tan B$ is positive. The angle $B$ is in the principal value branch of $\tan^{-1}$.

Now, substitute these back into the RHS:

RHS $= A + B = \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{4}{3}$.

We use the identity $\tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$. This identity is valid for $x > 0$, $y > 0$, and $xy < 1$.

Here, $x = \frac{5}{12}$ and $y = \frac{4}{3}$. Both are positive.

Check the product $xy$: $xy = \left(\frac{5}{12}\right)\left(\frac{4}{3}\right) = \frac{20}{36} = \frac{5}{9}$.

Since $xy = \frac{5}{9} < 1$, the identity is applicable.

Calculate the argument $\frac{x+y}{1-xy}$:

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 3.

So, applying the identity:

This is equal to the Left Hand Side (LHS).

Thus, $\tan^{-1} \frac{63}{16} = \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}$ is 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:

Consider the Right Hand Side (RHS) of the equation:

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

Let $x = \tan^2 \theta$ for some angle $\theta$.

Since $x \in [0, 1]$, $\sqrt{x} \in [0, 1]$. This implies $\tan \theta \in [0, 1]$.

Thus, $\theta \in [0, \frac{\pi}{4}]$. Consequently, $2\theta \in [0, \frac{\pi}{2}]$.

Substitute $x = \tan^2 \theta$ into the expression inside the inverse cosine function:

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

We use the double angle identity for cosine: $\cos 2\theta = \frac{1-\tan^2 \theta}{1+\tan^2 \theta}$.

So, $\frac{1-x}{1+x} = \cos 2\theta$.

Now, substitute this back into the RHS:

RHS $= \frac{1}{2} \cos^{-1} (\cos 2\theta)$

Since $2\theta \in [0, \frac{\pi}{2}]$, and in the interval $[0, \pi]$, $\cos^{-1}(\cos y) = y$, we have:

RHS $= \frac{1}{2} (2\theta) = \theta$

From our initial substitution, $x = \tan^2 \theta$, taking the square root of both sides gives $\sqrt{x} = \tan \theta$.

Applying the inverse tangent function to both sides, we get $\theta = \tan^{-1} \sqrt{x}$.

Therefore, RHS $= \tan^{-1} \sqrt{x}$.

This is the Left Hand Side (LHS) of the given equation.

Thus, LHS = RHS.

The identity is proven for $x \in [0, 1]$.

Given:

The expression is $\cot^{-1} \left( \frac{\sqrt{1+\sin\;x}\;+\;\sqrt{1-sin\;x}}{\sqrt{1+sin\;x}\;-\;\sqrt{1-sin\;x}} \right)$ for $x \in \left( 0,\frac{\pi}{4} \right)$.

To Prove:

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

Solution:

Consider the expression inside the inverse cotangent function:

$\frac{\sqrt{1+\sin\;x}\;+\;\sqrt{1-sin\;x}}{\sqrt{1+sin\;x}\;-\;\sqrt{1-sin\;x}}$

We know that $1 = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$.

Therefore,

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

And

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

Since $x \in \left( 0, \frac{\pi}{4} \right)$, it implies $\frac{x}{2} \in \left( 0, \frac{\pi}{8} \right)$.

For $\frac{x}{2} \in \left( 0, \frac{\pi}{8} \right)$, both $\sin \frac{x}{2}$ and $\cos \frac{x}{2}$ are positive. Also, $\cos \theta > \sin \theta$ for $\theta \in \left( 0, \frac{\pi}{4} \right)$, so $\cos \frac{x}{2} > \sin \frac{x}{2}$.

Thus, we can take the square roots directly:

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

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

Substitute these into the given expression:

$L.H.S. = \cot^{-1} \left( \frac{(\sin \frac{x}{2} + \cos \frac{x}{2}) + (\cos \frac{x}{2} - \sin \frac{x}{2})}{(\sin \frac{x}{2} + \cos \frac{x}{2}) - (\cos \frac{x}{2} - \sin \frac{x}{2})} \right)$

Simplify the numerator and the denominator:

Numerator $= \sin \frac{x}{2} + \cos \frac{x}{2} + \cos \frac{x}{2} - \sin \frac{x}{2} = 2 \cos \frac{x}{2}$

Denominator $= \sin \frac{x}{2} + \cos \frac{x}{2} - \cos \frac{x}{2} + \sin \frac{x}{2} = 2 \sin \frac{x}{2}$

So, the expression becomes:

$\frac{2 \cos \frac{x}{2}}{2 \sin \frac{x}{2}} = \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} = \cot \frac{x}{2}$

Therefore,

$L.H.S. = \cot^{-1} \left( \cot \frac{x}{2} \right)$

Since $x \in \left( 0, \frac{\pi}{4} \right)$, we have $\frac{x}{2} \in \left( 0, \frac{\pi}{8} \right)$. The range of the principal value branch of $\cot^{-1}(y)$ is $(0, \pi)$.

Since $\frac{x}{2} \in \left( 0, \frac{\pi}{8} \right)$ is within the range $(0, \pi)$, we have $\cot^{-1} \left( \cot \frac{x}{2} \right) = \frac{x}{2}$.

$L.H.S. = \frac{x}{2}$

This is equal to the R.H.S.

Hence, the identity is proven.

Given:

The expression is $\tan^{-1} \left( \frac{\sqrt{1+x}\;-\;\sqrt{1-x}}{\sqrt{1+x}\;+\;\sqrt{1-x}} \right)$ for $-\frac{1}{\sqrt{2}} \le x \le 1$.

To Prove:

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

Solution:

Let the given expression be $L.H.S.$

$L.H.S. = \tan^{-1} \left( \frac{\sqrt{1+x}\;-\;\sqrt{1-x}}{\sqrt{1+x}\;+\;\sqrt{1-x}} \right)$

Using the hint, let $x = \cos 2\theta$.

Since $-\frac{1}{\sqrt{2}} \le x \le 1$, we have $-\frac{1}{\sqrt{2}} \le \cos 2\theta \le 1$.

The principal value branch of $\cos^{-1} y$ is $[0, \pi]$. Applying $\cos^{-1}$ to the inequality (note that $\cos^{-1}$ is a decreasing function), we get:

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

$0 \le 2\theta \le \frac{3\pi}{4}$

Dividing by 2, we get the range for $\theta$:

$0 \le \theta \le \frac{3\pi}{8}$

Now substitute $x = \cos 2\theta$ into the expression:

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

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

For the range $0 \le \theta \le \frac{3\pi}{8}$, $\theta$ is in the first quadrant, where $\cos \theta \ge 0$ and $\sin \theta \ge 0$.

So, $|\cos \theta| = \cos \theta$ and $|\sin \theta| = \sin \theta$.

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

$\frac{\sqrt{2}\cos \theta - \sqrt{2}\sin \theta}{\sqrt{2}\cos \theta + \sqrt{2}\sin \theta} = \frac{\sqrt{2}(\cos \theta - \sin \theta)}{\sqrt{2}(\cos \theta + \sin \theta)} = \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}$

Divide the numerator and the denominator by $\cos \theta$ (since $\cos \theta \ne 0$ for $0 \le \theta \le \frac{3\pi}{8}$):

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

Using the tangent addition formula $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$ and knowing that $\tan \frac{\pi}{4} = 1$, we have:

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

Now substitute this back into the $L.H.S.$:

$L.H.S. = \tan^{-1} \left(\tan \left(\frac{\pi}{4} - \theta\right)\right)$

We need to check if the argument $\frac{\pi}{4} - \theta$ is in the principal value range of $\tan^{-1}$, which is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

From $0 \le \theta \le \frac{3\pi}{8}$, we have $-\frac{3\pi}{8} \le -\theta \le 0$.

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

$\frac{\pi}{4} - \frac{3\pi}{8} \le \frac{\pi}{4} - \theta \le \frac{\pi}{4} + 0$

$\frac{2\pi - 3\pi}{8} \le \frac{\pi}{4} - \theta \le \frac{\pi}{4}$

$-\frac{\pi}{8} \le \frac{\pi}{4} - \theta \le \frac{\pi}{4}$

The interval $[-\frac{\pi}{8}, \frac{\pi}{4}]$ is within the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.

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

$L.H.S. = \frac{\pi}{4} - \theta$

Now, we substitute back $\theta$ in terms of $x$. Since $x = \cos 2\theta$, we have $2\theta = \cos^{-1} x$.

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

$L.H.S. = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x$

This is equal to the $R.H.S.$

Hence, the identity is proven.

Given:

The equation is $2\tan^{-1}(\cos x) = \tan^{-1}(2 \text{ cosec } x)$.

To Solve:

Find the value(s) of $x$ that satisfy the given equation.

Solution:

The given equation is:

We use the formula $2\tan^{-1} y = \tan^{-1} \left(\frac{2y}{1-y^2}\right)$, provided $|y| < 1$. Here $y = \cos x$.

Applying this formula to the left-hand side ($L.H.S.$) of equation (i):

$L.H.S. = \tan^{-1} \left(\frac{2\cos x}{1-\cos^2 x}\right)$

Using the identity $1-\cos^2 x = \sin^2 x$, we get:

$L.H.S. = \tan^{-1} \left(\frac{2\cos x}{\sin^2 x}\right)$

Rewrite the term inside the $\tan^{-1}$ function:

$\frac{2\cos x}{\sin^2 x} = \frac{2\cos x}{\sin x \cdot \sin x} = 2 \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} = 2 \cot x \cdot \text{cosec } x$

So, the equation becomes:

$\tan^{-1}(2 \cot x \text{ cosec } x) = \tan^{-1}(2 \text{ cosec } x)$

If $\tan^{-1} A = \tan^{-1} B$, then $A=B$. Therefore, we can equate the arguments, provided they are defined:

$2 \cot x \text{ cosec } x = 2 \text{ cosec } x$

Subtract $2 \text{ cosec } x$ from both sides:

$2 \cot x \text{ cosec } x - 2 \text{ cosec } x = 0$

Factor out $2 \text{ cosec } x$:

$2 \text{ cosec } x (\cot x - 1) = 0$

This equation is satisfied if either $2 \text{ cosec } x = 0$ or $\cot x - 1 = 0$.

Case 1: $2 \text{ cosec } x = 0$

$\text{cosec } x = 0$

Since $\text{cosec } x = \frac{1}{\sin x}$, this means $\frac{1}{\sin x} = 0$, which is impossible for any real value of $x$. So, there are no solutions from this case.

Case 2: $\cot x - 1 = 0$

$\cot x = 1$

The general solution for $\cot x = 1$ is $x = n\pi + \frac{\pi}{4}$, where $n$ is an integer ($n \in \mathbb{Z}$).

We need to verify if these solutions are valid for the original equation. The formula for $2\tan^{-1}y$ requires $|y| < 1$, i.e., $|\cos x| < 1$. Also, $\tan^{-1}(2 \text{ cosec } x)$ requires $\text{cosec } x$ to be defined, which means $\sin x \ne 0$.

For $x = n\pi + \frac{\pi}{4}$:

$\cos x = \cos(n\pi + \frac{\pi}{4}) = \begin{cases} \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} & \text{if } n \text{ is even} \\ \cos(\pi + \frac{\pi}{4}) = -\cos(\frac{\pi}{4}) = -\frac{1}{\sqrt{2}} & \text{if } n \text{ is odd} \end{cases}$

In both cases, $|\cos x| = \left|\pm\frac{1}{\sqrt{2}}\right| = \frac{1}{\sqrt{2}}$, which is less than 1. So, the condition $|\cos x| < 1$ is satisfied.

$\sin x = \sin(n\pi + \frac{\pi}{4}) = \begin{cases} \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} & \text{if } n \text{ is even} \\ \sin(\pi + \frac{\pi}{4}) = -\sin(\frac{\pi}{4}) = -\frac{1}{\sqrt{2}} & \text{if } n \text{ is odd} \end{cases}$

In both cases, $\sin x \ne 0$, so $\text{cosec } x$ is defined.

Thus, the solutions $x = n\pi + \frac{\pi}{4}$ for $n \in \mathbb{Z}$ are valid.

The solution to the equation is $x = n\pi + \frac{\pi}{4}$, where $n$ is any integer.

Given:

The equation is $\tan^{-1} \left( \frac{1 - x}{1 + x} \right) = \frac{1}{2} \tan^{-1} x$, with the condition $x > 0$.

To Solve:

Find the value of $x$ that satisfies the given equation.

Solution:

The given equation is:

We use the identity $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$. Setting $A=1$ and $B=x$, the left-hand side ($L.H.S.$) can be written as:

$L.H.S. = \tan^{-1} 1 - \tan^{-1} x$

We know that $\tan^{-1} 1 = \frac{\pi}{4}$.

$L.H.S. = \frac{\pi}{4} - \tan^{-1} x$

Substitute this into equation (i):

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

Add $\tan^{-1} x$ to both sides:

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

Combine the terms on the right-hand side:

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

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

Multiply both sides by $\frac{2}{3}$ to solve for $\tan^{-1} x$:

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

$\tan^{-1} x = \frac{2\pi}{12}$

$\tan^{-1} x = \frac{\pi}{6}$

Now, take the tangent of both sides to find $x$:

$x = \tan \left(\frac{\pi}{6}\right)$

We know that $\tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$.

We need to check if this solution satisfies the given condition $x > 0$. Since $\frac{1}{\sqrt{3}} > 0$, the solution is valid.

The identity $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$ is valid when $AB > -1$. Here $A=1$ and $B=x = \frac{1}{\sqrt{3}}$. $AB = 1 \times \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}}$. Since $\frac{1}{\sqrt{3}} > -1$, the identity used is valid for this solution.

The solution to the equation is $x = \frac{1}{\sqrt{3}}$.

Given:

The expression $\sin (\tan^{-1} x)$, with the condition $|x| < 1$. The options are provided.

To Find:

The expression that is equal to $\sin (\tan^{-1} x)$ from the given options.

Solution:

Let $\theta = \tan^{-1} x$.

By the definition of the inverse tangent function, this means $\tan \theta = x$.

Since $|x| < 1$, the principal value of $\tan^{-1} x$ lies in the interval $(-\frac{\pi}{4}, \frac{\pi}{4})$. Thus, $\theta \in (-\frac{\pi}{4}, \frac{\pi}{4})$.

We need to find $\sin \theta$ where $\tan \theta = x$.

We can relate $\sin \theta$ and $\tan \theta$ using trigonometric identities.

Substitute $\tan \theta = x$ into the identity:

$\sin^2 \theta = \frac{x^2}{1 + x^2}$

Taking the square root of both sides:

$\sin \theta = \pm \sqrt{\frac{x^2}{1 + x^2}} = \pm \frac{|x|}{\sqrt{1 + x^2}}$

We need to determine the correct sign. Since $\theta \in (-\frac{\pi}{4}, \frac{\pi}{4})$, $\theta$ is in the first or fourth quadrant.

• If $x > 0$, then $\tan \theta = x > 0$, which implies $\theta \in (0, \frac{\pi}{4})$. In the first quadrant, $\sin \theta$ is positive. So, $\sin \theta = \frac{|x|}{\sqrt{1 + x^2}} = \frac{x}{\sqrt{1 + x^2}}$ (since $x>0, |x|=x$).
• If $x < 0$, then $\tan \theta = x < 0$, which implies $\theta \in (-\frac{\pi}{4}, 0)$. In the fourth quadrant, $\sin \theta$ is negative. So, $\sin \theta = -\frac{|x|}{\sqrt{1 + x^2}}$. Since $x < 0$, $|x| = -x$. Therefore, $\sin \theta = -\frac{-x}{\sqrt{1 + x^2}} = \frac{x}{\sqrt{1 + x^2}}$.
• If $x = 0$, then $\tan \theta = 0$, which implies $\theta = 0$. $\sin \theta = \sin 0 = 0$. The formula $\frac{x}{\sqrt{1+x^2}} = \frac{0}{\sqrt{1+0^2}} = 0$, which matches.

In all cases where $|x| < 1$, we find that $\sin \theta = \frac{x}{\sqrt{1 + x^2}}$.

Substituting back $\theta = \tan^{-1} x$:

Comparing this result with the given options:

(A) $\frac{x}{\sqrt{1-x^{2}}}$

(B) $\frac{1}{\sqrt{1-x^{2}}}$

(C) $\frac{1}{\sqrt{1+x^{2}}}$

(D) $\frac{x}{\sqrt{1+x^{2}}}$

The result matches option (D).

The correct option is (D).

Given:

The equation is $\sin^{-1} (1 – x) – 2 \sin^{–1} x = \frac{\pi}{2}$.

To Solve:

Find the value(s) of $x$ that satisfy the given equation.

Solution:

The given equation is:

First, let's consider the domain of the inverse sine function. For $\sin^{-1} z$ to be defined, we must have $-1 \le z \le 1$.

For $\sin^{-1}(1-x)$: $-1 \le 1-x \le 1$. Subtracting 1 from all parts gives $-2 \le -x \le 0$. Multiplying by $-1$ and reversing the inequalities gives $0 \le x \le 2$.

For $\sin^{-1}x$: $-1 \le x \le 1$.

For both inverse sine functions to be defined, $x$ must be in the intersection of the two domains: $[0, 2] \cap [-1, 1] = [0, 1]$. So, any valid solution must be in the interval $[0, 1]$.

Now, let's solve the equation. Rearrange equation (i):

$\sin^{-1} (1 – x) = \frac{\pi}{2} + 2 \sin^{–1} x$

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

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

Taking the sine of both sides:

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

Using the identity $\sin(\sin^{-1} z) = z$ (provided $z$ is in the domain of $\sin^{-1}$) and $\sin\left(\frac{\pi}{2} + \theta\right) = \cos \theta$, we get:

$1 - x = \cos(2y)$

Using the double angle identity for cosine, $\cos(2y) = 1 - 2\sin^2 y$. Since $y = \sin^{-1} x$, we have $\sin y = x$. Substituting this into the identity:

$\cos(2y) = 1 - 2x^2$

So, the equation becomes:

$1 - x = 1 - 2x^2$

Subtract 1 from both sides:

$-x = -2x^2$

Rearrange into a quadratic equation:

$2x^2 - x = 0$

Factor out $x$:

$x(2x - 1) = 0$

This gives two possible solutions: $x = 0$ or $2x - 1 = 0 \implies x = \frac{1}{2}$.

These are potential solutions obtained from the algebraic simplification. We must check if they satisfy the original equation (i) and lie within the domain $[0, 1]$. Both $x=0$ and $x=\frac{1}{2}$ are in the domain $[0, 1]$.

Verification:

Check $x = 0$ in equation (i):

$L.H.S. = \sin^{-1} (1 – 0) – 2 \sin^{–1} 0 = \sin^{-1} 1 – 2 \sin^{–1} 0 = \frac{\pi}{2} – 2(0) = \frac{\pi}{2}$

$R.H.S. = \frac{\pi}{2}$

Since $L.H.S. = R.H.S.$, $x = 0$ is a solution.

Check $x = \frac{1}{2}$ in equation (i):

$L.H.S. = \sin^{-1} \left(1 – \frac{1}{2}\right) – 2 \sin^{–1} \frac{1}{2} = \sin^{-1} \frac{1}{2} – 2 \sin^{–1} \frac{1}{2}$

Using the principal values, $\sin^{-1} \frac{1}{2} = \frac{\pi}{6}$.

$L.H.S. = \frac{\pi}{6} – 2 \left(\frac{\pi}{6}\right) = \frac{\pi}{6} – \frac{\pi}{3} = \frac{\pi - 2\pi}{6} = -\frac{\pi}{6}$

$R.H.S. = \frac{\pi}{2}$

Since $L.H.S. \ne R.H.S.$, $x = \frac{1}{2}$ is not a solution.

Thus, the only solution to the equation is $x = 0$.

The solution is $x = 0$, which corresponds to option (C).