Class 12-NCERT Solutions-Chapter-02-Inverse Trigonometric Functions-Ex 2.2

Solution:

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

Consider the Right Hand Side (RHS):

\[ \begin{aligned} \text{RHS} &= \sin^{-1} (3x - 4x^3) \\ &= \sin^{-1} (3\sin \theta - 4\sin^3 \theta) \\ &= \sin^{-1} (\sin 3\theta) \quad [\text{Since } \sin 3\theta = 3\sin \theta - 4\sin^3 \theta] \\ &= 3\theta \\ &= 3\sin^{-1} x \\ &= \text{LHS} \end{aligned} \]

Therefore, \( 3\sin^{-1} x = \sin^{-1} (3x - 4x^3) \).

Solution:

Let \( x = \cos \theta \). Then \( \theta = \cos^{-1} x \).

Consider the Right Hand Side (RHS):

\[ \begin{aligned} \text{RHS} &= \cos^{-1} (4x^3 - 3x) \\ &= \cos^{-1} (4\cos^3 \theta - 3\cos \theta) \\ &= \cos^{-1} (\cos 3\theta) \quad [\text{Since } \cos 3\theta = 4\cos^3 \theta - 3\cos \theta] \\ &= 3\theta \\ &= 3\cos^{-1} x \\ &= \text{LHS} \end{aligned} \]

Therefore, \( 3\cos^{-1} x = \cos^{-1} (4x^3 - 3x) \).

Solution:

Let \( x = \tan \theta \). Then \( \theta = \tan^{-1} x \).

Substituting \( x = \tan \theta \) in the given expression:

\[ \begin{aligned} \tan^{-1} \frac{\sqrt{1+x^2}-1}{x} &= \tan^{-1} \frac{\sqrt{1+\tan^2 \theta}-1}{\tan \theta} \\ &= \tan^{-1} \frac{\sec \theta - 1}{\tan \theta} \\ &= \tan^{-1} \left( \frac{\frac{1}{\cos \theta} - 1}{\frac{\sin \theta}{\cos \theta}} \right) \\ &= \tan^{-1} \left( \frac{1 - \cos \theta}{\sin \theta} \right) \end{aligned} \]

Using half-angle formulas \( 1 - \cos \theta = 2\sin^2 \frac{\theta}{2} \) and \( \sin \theta = 2\sin \frac{\theta}{2} \cos \frac{\theta}{2} \):

\[ \begin{aligned} &= \tan^{-1} \left( \frac{2\sin^2 \frac{\theta}{2}}{2\sin \frac{\theta}{2} \cos \frac{\theta}{2}} \right) \\ &= \tan^{-1} \left( \tan \frac{\theta}{2} \right) \\ &= \frac{\theta}{2} \\ &= \frac{1}{2} \tan^{-1} x \end{aligned} \]

Solution:

Using the identities \( 1 - \cos x = 2\sin^2 \frac{x}{2} \) and \( 1 + \cos x = 2\cos^2 \frac{x}{2} \):

\[ \begin{aligned} \tan^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}} &= \tan^{-1} \sqrt{\frac{2\sin^2 \frac{x}{2}}{2\cos^2 \frac{x}{2}}} \\ &= \tan^{-1} \sqrt{\tan^2 \frac{x}{2}} \\ &= \tan^{-1} \left( \tan \frac{x}{2} \right) \\ &= \frac{x}{2} \end{aligned} \]

Solution:

Divide the numerator and denominator inside the parenthesis by \( \cos x \):

\[ \begin{aligned} \tan^{-1} \left( \frac{1 - \frac{\sin x}{\cos x}}{1 + \frac{\sin x}{\cos x}} \right) &= \tan^{-1} \left( \frac{1 - \tan x}{1 + \tan x} \right) \end{aligned} \]

We know that \( \tan \frac{\pi}{4} = 1 \). So, the expression becomes:

\[ \begin{aligned} &= \tan^{-1} \left( \frac{\tan \frac{\pi}{4} - \tan x}{1 + \tan \frac{\pi}{4} \tan x} \right) \\ &= \tan^{-1} \left( \tan \left( \frac{\pi}{4} - x \right) \right) \quad [\text{Using } \tan(A-B) \text{ formula}] \\ &= \frac{\pi}{4} - x \end{aligned} \]

Solution:

Let \( x = a \sin \theta \). Then \( \sin \theta = \frac{x}{a} \) or \( \theta = \sin^{-1} \frac{x}{a} \).

Substitute this into the expression:

\[ \begin{aligned} \tan^{-1} \frac{a \sin \theta}{\sqrt{a^2 - a^2 \sin^2 \theta}} &= \tan^{-1} \frac{a \sin \theta}{a \sqrt{1 - \sin^2 \theta}} \\ &= \tan^{-1} \frac{\sin \theta}{\cos \theta} \\ &= \tan^{-1} (\tan \theta) \\ &= \theta \\ &= \sin^{-1} \frac{x}{a} \end{aligned} \]

Solution:

Let \( x = a \tan \theta \). Then \( \tan \theta = \frac{x}{a} \) or \( \theta = \tan^{-1} \frac{x}{a} \).

Substitute \( x = a \tan \theta \) into the expression:

\[ \begin{aligned} \tan^{-1} \left( \frac{3a^2(a \tan \theta) - (a \tan \theta)^3}{a^3 - 3a(a \tan \theta)^2} \right) &= \tan^{-1} \left( \frac{3a^3 \tan \theta - a^3 \tan^3 \theta}{a^3 - 3a^3 \tan^2 \theta} \right) \\ &= \tan^{-1} \left( \frac{a^3(3 \tan \theta - \tan^3 \theta)}{a^3(1 - 3 \tan^2 \theta)} \right) \\ &= \tan^{-1} \left( \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \right) \\ &= \tan^{-1} (\tan 3\theta) \quad [\text{Using } \tan 3A \text{ formula}] \\ &= 3\theta \\ &= 3 \tan^{-1} \frac{x}{a} \end{aligned} \]

Solution:

First, find the principal value of \( \sin^{-1} \frac{1}{2} \).

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

Now substitute this back into the expression:

\[ \begin{aligned} \tan^{-1} \left[ 2\cos \left( 2 \cdot \frac{\pi}{6} \right) \right] &= \tan^{-1} \left[ 2\cos \left( \frac{\pi}{3} \right) \right] \\ &= \tan^{-1} \left[ 2 \left( \frac{1}{2} \right) \right] \quad [\text{Since } \cos \frac{\pi}{3} = \frac{1}{2}] \\ &= \tan^{-1} (1) \\ &= \frac{\pi}{4} \end{aligned} \]

Solution:

We know the standard inverse trigonometric identities:

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

Substitute these into the expression:

\[ \begin{aligned} \tan \frac{1}{2} [ 2\tan^{-1} x + 2\tan^{-1} y ] &= \tan \frac{1}{2} [ 2(\tan^{-1} x + \tan^{-1} y) ] \\ &= \tan (\tan^{-1} x + \tan^{-1} y) \\ &= \tan \left( \tan^{-1} \frac{x+y}{1-xy} \right) \\ &= \frac{x+y}{1-xy} \end{aligned} \]

Solution:

We know that \( \sin^{-1}(\sin x) = x \) only if \( x \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \).

Here, \( \frac{2\pi}{3} \notin \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \). So we must adjust the angle.

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

Therefore:

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

Solution:

We know that \( \tan^{-1}(\tan x) = x \) only if \( x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \).

Here, \( \frac{3\pi}{4} \notin \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \).

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

Therefore:

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

Solution:

Convert both inverse functions to \( \tan^{-1} \):

1. Let \( \sin^{-1} \frac{3}{5} = x \). Then \( \sin x = \frac{3}{5} \). Thus \( \tan x = \frac{3}{4} \), so \( \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{3}{4} \).

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

Now substitute these:

\[ \begin{aligned} \tan \left( \tan^{-1} \frac{3}{4} + \tan^{-1} \frac{2}{3} \right) &= \tan \left( \tan^{-1} \frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \cdot \frac{2}{3}} \right) \\ &= \frac{\frac{9+8}{12}}{1 - \frac{6}{12}} \\ &= \frac{\frac{17}{12}}{\frac{6}{12}} \\ &= \frac{17}{6} \end{aligned} \]

Answer: (B)

Solution:

The range of \( \cos^{-1} \) is \( [0, \pi] \). Since \( \frac{7\pi}{6} > \pi \), it is not the principal value.

\[ \cos \frac{7\pi}{6} = \cos \left( 2\pi - \frac{5\pi}{6} \right) = \cos \frac{5\pi}{6} \]

Also, \( \frac{5\pi}{6} \in [0, \pi] \).

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

Answer: (D)

Solution:

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

Substitute this into the expression:

\[ \begin{aligned} \sin \left( \frac{\pi}{3} - \left( -\frac{\pi}{6} \right) \right) &= \sin \left( \frac{\pi}{3} + \frac{\pi}{6} \right) \\ &= \sin \left( \frac{2\pi + \pi}{6} \right) \\ &= \sin \left( \frac{3\pi}{6} \right) \\ &= \sin \left( \frac{\pi}{2} \right) \\ &= 1 \end{aligned} \]

Answer: (B)

Solution:

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

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

Subtract the values:

\[ \frac{\pi}{3} - \frac{5\pi}{6} = \frac{2\pi - 5\pi}{6} = \frac{-3\pi}{6} = -\frac{\pi}{2} \]