Class 12-NCERT Solutions-Chapter-02-Inverse Trigonometric Functions-Mis

Solution:

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

Here, \( \frac{13\pi}{6} > \pi \), so we simplify the angle.

\[ \cos \frac{13\pi}{6} = \cos \left( 2\pi + \frac{\pi}{6} \right) = \cos \frac{\pi}{6} \]

Since \( \frac{\pi}{6} \in [0, \pi] \):

\[ \cos^{-1} \left( \cos \frac{13\pi}{6} \right) = \cos^{-1} \left( \cos \frac{\pi}{6} \right) = \frac{\pi}{6} \]

Solution:

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

Here, \( \frac{7\pi}{6} \notin \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \).

\[ \tan \frac{7\pi}{6} = \tan \left( \pi + \frac{\pi}{6} \right) = \tan \frac{\pi}{6} \]

Since \( \frac{\pi}{6} \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \):

\[ \tan^{-1} \left( \tan \frac{7\pi}{6} \right) = \tan^{-1} \left( \tan \frac{\pi}{6} \right) = \frac{\pi}{6} \]

Solution:

Let \( \sin^{-1} \frac{3}{5} = x \). Then \( \sin x = \frac{3}{5} \).

Using Pythagoras theorem, \( \cos x = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5} \).

Therefore, \( \tan x = \frac{3}{4} \Rightarrow x = \tan^{-1} \frac{3}{4} \).

Now, consider LHS:

\[ \begin{aligned} 2\sin^{-1} \frac{3}{5} &= 2\tan^{-1} \frac{3}{4} \\ &= \tan^{-1} \left( \frac{2 \cdot \frac{3}{4}}{1 - \left(\frac{3}{4}\right)^2} \right) \\ &= \tan^{-1} \left( \frac{\frac{3}{2}}{1 - \frac{9}{16}} \right) \\ &= \tan^{-1} \left( \frac{\frac{3}{2}}{\frac{7}{16}} \right) \\ &= \tan^{-1} \left( \frac{3}{2} \times \frac{16}{7} \right) \\ &= \tan^{-1} \frac{24}{7} = \text{RHS} \end{aligned} \]

Solution:

Convert sine inverse terms to tangent inverse:

1. Let \( \sin^{-1} \frac{8}{17} = x \Rightarrow \tan x = \frac{8}{15} \Rightarrow x = \tan^{-1} \frac{8}{15} \).

2. Let \( \sin^{-1} \frac{3}{5} = y \Rightarrow \tan y = \frac{3}{4} \Rightarrow y = \tan^{-1} \frac{3}{4} \).

LHS becomes:

\[ \begin{aligned} \tan^{-1} \frac{8}{15} + \tan^{-1} \frac{3}{4} &= \tan^{-1} \left( \frac{\frac{8}{15} + \frac{3}{4}}{1 - \frac{8}{15} \cdot \frac{3}{4}} \right) \\ &= \tan^{-1} \left( \frac{\frac{32+45}{60}}{1 - \frac{24}{60}} \right) \\ &= \tan^{-1} \left( \frac{77}{60-24} \right) \\ &= \tan^{-1} \frac{77}{36} = \text{RHS} \end{aligned} \]

Solution:

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

Here, \( x = \frac{4}{5} \) and \( y = \frac{12}{13} \).

\[ \begin{aligned} \text{LHS} &= \cos^{-1} \left( \frac{4}{5} \cdot \frac{12}{13} - \sqrt{1 - \left(\frac{4}{5}\right)^2} \sqrt{1 - \left(\frac{12}{13}\right)^2} \right) \\ &= \cos^{-1} \left( \frac{48}{65} - \sqrt{\frac{9}{25}} \sqrt{\frac{25}{169}} \right) \\ &= \cos^{-1} \left( \frac{48}{65} - \frac{3}{5} \cdot \frac{5}{13} \right) \\ &= \cos^{-1} \left( \frac{48}{65} - \frac{15}{65} \right) \\ &= \cos^{-1} \frac{33}{65} = \text{RHS} \end{aligned} \]

Solution:

Convert \( \cos^{-1} \frac{12}{13} \) to \( \sin^{-1} \).

Let \( \cos^{-1} \frac{12}{13} = x \Rightarrow \cos x = \frac{12}{13} \). Then \( \sin x = \frac{5}{13} \), so \( x = \sin^{-1} \frac{5}{13} \).

Now use the formula \( \sin^{-1} x + \sin^{-1} y = \sin^{-1} (x\sqrt{1-y^2} + y\sqrt{1-x^2}) \).

Here \( x = \frac{5}{13} \) and \( y = \frac{3}{5} \).

\[ \begin{aligned} \text{LHS} &= \sin^{-1} \frac{5}{13} + \sin^{-1} \frac{3}{5} \\ &= \sin^{-1} \left( \frac{5}{13}\sqrt{1-\frac{9}{25}} + \frac{3}{5}\sqrt{1-\frac{25}{169}} \right) \\ &= \sin^{-1} \left( \frac{5}{13} \cdot \frac{4}{5} + \frac{3}{5} \cdot \frac{12}{13} \right) \\ &= \sin^{-1} \left( \frac{20}{65} + \frac{36}{65} \right) \\ &= \sin^{-1} \frac{56}{65} = \text{RHS} \end{aligned} \]

Solution:

Consider the RHS. Convert both terms to \( \tan^{-1} \).

1. \( \sin^{-1} \frac{5}{13} = \tan^{-1} \frac{5}{12} \).

2. \( \cos^{-1} \frac{3}{5} = \tan^{-1} \frac{4}{3} \).

\[ \begin{aligned} \text{RHS} &= \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{4}{3} \\ &= \tan^{-1} \left( \frac{\frac{5}{12} + \frac{4}{3}}{1 - \frac{5}{12} \cdot \frac{4}{3}} \right) \\ &= \tan^{-1} \left( \frac{\frac{5+16}{12}}{1 - \frac{20}{36}} \right) \\ &= \tan^{-1} \left( \frac{\frac{21}{12}}{\frac{16}{36}} \right) \\ &= \tan^{-1} \left( \frac{7}{4} \cdot \frac{9}{4} \right) \\ &= \tan^{-1} \frac{63}{16} = \text{LHS} \end{aligned} \]

Solution:

Let \( \sqrt{x} = \tan \theta \). Then \( x = \tan^2 \theta \). So, \( \theta = \tan^{-1} \sqrt{x} \).

Substitute \( x \) in the RHS:

\[ \begin{aligned} \text{RHS} &= \frac{1}{2} \cos^{-1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) \\ &= \frac{1}{2} \cos^{-1} (\cos 2\theta) \quad [\text{Using } \cos 2\theta \text{ formula}] \\ &= \frac{1}{2} (2\theta) \\ &= \theta \\ &= \tan^{-1} \sqrt{x} = \text{LHS} \end{aligned} \]

Solution:

We use the identity \( 1 \pm \sin x = \left( \cos \frac{x}{2} \pm \sin \frac{x}{2} \right)^2 \).

Since \( x \in \left( 0, \frac{\pi}{4} \right) \), \( \frac{x}{2} \in \left( 0, \frac{\pi}{8} \right) \), so \( \cos \frac{x}{2} > \sin \frac{x}{2} > 0 \).

Therefore, \( \sqrt{1+\sin x} = \cos \frac{x}{2} + \sin \frac{x}{2} \) and \( \sqrt{1-\sin x} = \cos \frac{x}{2} - \sin \frac{x}{2} \).

Substitute these into the expression:

\[ \begin{aligned} \text{LHS} &= \cot^{-1} \left( \frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) + (\cos \frac{x}{2} - \sin \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2}) - (\cos \frac{x}{2} - \sin \frac{x}{2})} \right) \\ &= \cot^{-1} \left( \frac{2\cos \frac{x}{2}}{2\sin \frac{x}{2}} \right) \\ &= \cot^{-1} \left( \cot \frac{x}{2} \right) \\ &= \frac{x}{2} = \text{RHS} \end{aligned} \]

Solution:

Put \( x = \cos 2\theta \). Then \( 2\theta = \cos^{-1} x \Rightarrow \theta = \frac{1}{2}\cos^{-1} x \).

\[ \begin{aligned} \text{LHS} &= \tan^{-1} \left( \frac{\sqrt{1+\cos 2\theta} - \sqrt{1-\cos 2\theta}}{\sqrt{1+\cos 2\theta} + \sqrt{1-\cos 2\theta}} \right) \\ &= \tan^{-1} \left( \frac{\sqrt{2\cos^2 \theta} - \sqrt{2\sin^2 \theta}}{\sqrt{2\cos^2 \theta} + \sqrt{2\sin^2 \theta}} \right) \\ &= \tan^{-1} \left( \frac{\sqrt{2}\cos \theta - \sqrt{2}\sin \theta}{\sqrt{2}\cos \theta + \sqrt{2}\sin \theta} \right) \\ &= \tan^{-1} \left( \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta} \right) \\ &= \tan^{-1} \left( \tan \left( \frac{\pi}{4} - \theta \right) \right) \\ &= \frac{\pi}{4} - \theta \\ &= \frac{\pi}{4} - \frac{1}{2}\cos^{-1} x = \text{RHS} \end{aligned} \]

Solution:

Using the identity \( 2\tan^{-1} A = \tan^{-1} \frac{2A}{1-A^2} \) on LHS:

\[ \begin{aligned} \tan^{-1} \left( \frac{2\cos x}{1-\cos^2 x} \right) &= \tan^{-1} (2\text{cosec } x) \\ \Rightarrow \frac{2\cos x}{\sin^2 x} &= \frac{2}{\sin x} \\ \Rightarrow \frac{\cos x}{\sin x} &= 1 \quad (\text{Since } \sin x \neq 0) \\ \Rightarrow \cot x &= 1 \\ \Rightarrow x &= \frac{\pi}{4} \end{aligned} \]

Solution:

We can write \( \tan^{-1} \frac{1-x}{1+x} \) as \( \tan^{-1} 1 - \tan^{-1} x \).

\[ \begin{aligned} \tan^{-1} 1 - \tan^{-1} x &= \frac{1}{2}\tan^{-1} x \\ \frac{\pi}{4} - \tan^{-1} x &= \frac{1}{2}\tan^{-1} x \\ \frac{\pi}{4} &= \frac{3}{2}\tan^{-1} x \\ \tan^{-1} x &= \frac{2\pi}{12} = \frac{\pi}{6} \\ x &= \tan \frac{\pi}{6} \\ x &= \frac{1}{\sqrt{3}} \end{aligned} \]

Answer: (D)

Solution:

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

We need to find \( \sin \theta \).

Using a right-angled triangle, Perpendicular = \( x \), Base = 1.

Hypotenuse = \( \sqrt{1^2 + x^2} = \sqrt{1+x^2} \).

\[ \sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{x}{\sqrt{1+x^2}} \]

Therefore, the correct option is (D).

Answer: (C)

Solution:

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

The equation becomes:

\[ \begin{aligned} \sin^{-1}(1-x) - 2\theta &= \frac{\pi}{2} \\ \sin^{-1}(1-x) &= \frac{\pi}{2} + 2\theta \\ 1-x &= \sin \left( \frac{\pi}{2} + 2\theta \right) \\ 1-x &= \cos 2\theta \\ 1-x &= 1 - 2\sin^2 \theta \\ 1-x &= 1 - 2x^2 \\ 2x^2 - x &= 0 \\ x(2x-1) &= 0 \end{aligned} \]

This gives \( x = 0 \) or \( x = \frac{1}{2} \).

We must check these values in the original equation:

1. If \( x = \frac{1}{2} \): \( \sin^{-1}\frac{1}{2} - 2\sin^{-1}\frac{1}{2} = -\frac{\pi}{6} \neq \frac{\pi}{2} \). (Rejected)

2. If \( x = 0 \): \( \sin^{-1} 1 - 2\sin^{-1} 0 = \frac{\pi}{2} - 0 = \frac{\pi}{2} \). (Accepted)

Therefore, \( x = 0 \).