Here are the complete solutions for the first 14 questions of the Miscellaneous Exercise on NCERT Class 12 Maths Chapter 2 (Inverse Trigonometric Functions).
Table of Contents
Find the value of the following:
Question 1. $\cos^{-1} \left(\cos \frac{13\pi}{6}\right)$
Solution:
We know that the Principal Value Branch (PVB) of $\cos^{-1} x$ is $[0, \pi]$. Since $\frac{13\pi}{6}$ is outside this range, we rewrite the angle:
$$\frac{13\pi}{6} = 2\pi + \frac{\pi}{6}$$
Using the property $\cos(2\pi + \theta) = \cos \theta$:
$$\cos \left(\frac{13\pi}{6}\right) = \cos \left(2\pi + \frac{\pi}{6}\right) = \cos \frac{\pi}{6}$$
Now, $\frac{\pi}{6} \in [0, \pi]$.
$$\cos^{-1} \left(\cos \frac{13\pi}{6}\right) = \cos^{-1} \left(\cos \frac{\pi}{6}\right) = \frac{\pi}{6}$$
$$\text{Value} = \frac{\pi}{6}$$
Question 2. $\tan^{-1} \left(\tan \frac{7\pi}{6}\right)$
Solution:
The PVB of $\tan^{-1} x$ is $(-\pi/2, \pi/2)$. Since $\frac{7\pi}{6}$ is outside this range, we rewrite the angle:
$$\frac{7\pi}{6} = \pi + \frac{\pi}{6}$$
Using the property $\tan(\pi + \theta) = \tan \theta$:
$$\tan \left(\frac{7\pi}{6}\right) = \tan \left(\pi + \frac{\pi}{6}\right) = \tan \frac{\pi}{6}$$
Now, $\frac{\pi}{6} \in (-\pi/2, \pi/2)$.
$$\tan^{-1} \left(\tan \frac{7\pi}{6}\right) = \tan^{-1} \left(\tan \frac{\pi}{6}\right) = \frac{\pi}{6}$$
$$\text{Value} = \frac{\pi}{6}$$
Prove that:
Question 3. $2 \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{24}{7}$
Solution:
LHS: $2 \sin^{-1} \frac{3}{5}$
- Convert $\sin^{-1}$ to $\tan^{-1}$:Let $\sin^{-1} \frac{3}{5} = \theta$. Then $\sin \theta = 3/5$.Using a right triangle, the opposite side is 3, and the hypotenuse is 5.Adjacent side = $\sqrt{5^2 – 3^2} = \sqrt{25 – 9} = 4$.Therefore, $\tan \theta = 3/4$, and $\theta = \tan^{-1} \frac{3}{4}$.$$LHS = 2 \tan^{-1} \frac{3}{4}$$
- Apply $2 \tan^{-1} x$ identity:Use the identity $2 \tan^{-1} x = \tan^{-1} \left(\frac{2x}{1-x^2}\right)$, where $x = 3/4$.$$LHS = \tan^{-1} \left( \frac{2(3/4)}{1 – (3/4)^2} \right) = \tan^{-1} \left( \frac{3/2}{1 – 9/16} \right)$$$$LHS = \tan^{-1} \left( \frac{3/2}{7/16} \right) = \tan^{-1} \left( \frac{3}{2} \cdot \frac{16}{7} \right)$$$$LHS = \tan^{-1} \left( \frac{3 \cdot 8}{7} \right) = \tan^{-1} \frac{24}{7} = \mathbf{RHS}$$Hence, Proved.
Question 4. $\sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{77}{36}$
Solution:
We convert both $\sin^{-1}$ terms to $\tan^{-1}$ and then use the $\tan^{-1} x + \tan^{-1} y$ identity.
- Convert $\sin^{-1} \frac{8}{17}$:If $\sin \alpha = 8/17$, adjacent side $= \sqrt{17^2 – 8^2} = \sqrt{289 – 64} = 15$.$\sin^{-1} \frac{8}{17} = \tan^{-1} \frac{8}{15}$.
- Convert $\sin^{-1} \frac{3}{5}$:If $\sin \beta = 3/5$, adjacent side $= \sqrt{5^2 – 3^2} = 4$.$\sin^{-1} \frac{3}{5} = \tan^{-1} \frac{3}{4}$.
- Apply $\tan^{-1} x + \tan^{-1} y$ identity:$$LHS = \tan^{-1} \frac{8}{15} + \tan^{-1} \frac{3}{4}$$$$LHS = \tan^{-1} \left( \frac{\frac{8}{15} + \frac{3}{4}}{1 – \frac{8}{15} \cdot \frac{3}{4}} \right) = \tan^{-1} \left( \frac{\frac{32 + 45}{60}}{1 – \frac{24}{60}} \right)$$$$LHS = \tan^{-1} \left( \frac{77/60}{36/60} \right) = \tan^{-1} \frac{77}{36} = \mathbf{RHS}$$Hence, Proved.
Question 5. $\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1} \frac{33}{65}$
Solution:
We use the identity $\cos^{-1} x + \cos^{-1} y = \cos^{-1} (xy – \sqrt{1-x^2} \sqrt{1-y^2})$.
Here $x = 4/5$ and $y = 12/13$.
- Calculate $\sqrt{1-x^2}$ and $\sqrt{1-y^2}$:$$\sqrt{1-x^2} = \sqrt{1 – (4/5)^2} = \sqrt{1 – 16/25} = \sqrt{9/25} = 3/5$$$$\sqrt{1-y^2} = \sqrt{1 – (12/13)^2} = \sqrt{1 – 144/169} = \sqrt{25/169} = 5/13$$
- Apply the identity:$$LHS = \cos^{-1} \left( \left(\frac{4}{5}\right)\left(\frac{12}{13}\right) – \left(\frac{3}{5}\right)\left(\frac{5}{13}\right) \right)$$$$LHS = \cos^{-1} \left( \frac{48}{65} – \frac{15}{65} \right)$$$$LHS = \cos^{-1} \left( \frac{48 – 15}{65} \right) = \cos^{-1} \frac{33}{65} = \mathbf{RHS}$$Hence, Proved.
Question 6. $\cos^{-1} \frac{12}{13} + \sin^{-1} \frac{3}{5} = \sin^{-1} \frac{56}{65}$
Solution:
We convert the $\cos^{-1}$ term to $\sin^{-1}$ and then use the $\sin^{-1} x + \sin^{-1} y$ identity.
- Convert $\cos^{-1} \frac{12}{13}$ to $\sin^{-1}$:If $\cos \alpha = 12/13$, opposite side $= \sqrt{13^2 – 12^2} = 5$.$\cos^{-1} \frac{12}{13} = \sin^{-1} \frac{5}{13}$.
- Apply $\sin^{-1} x + \sin^{-1} y$ identity:Use $\sin^{-1} x + \sin^{-1} y = \sin^{-1} (x\sqrt{1-y^2} + y\sqrt{1-x^2})$, where $x = 5/13$ and $y = 3/5$.$$\sqrt{1-x^2} = \sqrt{1 – (5/13)^2} = 12/13$$$$\sqrt{1-y^2} = \sqrt{1 – (3/5)^2} = 4/5$$$$LHS = \sin^{-1} \left( \frac{5}{13} \cdot \frac{4}{5} + \frac{3}{5} \cdot \frac{12}{13} \right)$$$$LHS = \sin^{-1} \left( \frac{20}{65} + \frac{36}{65} \right)$$$$LHS = \sin^{-1} \left( \frac{20 + 36}{65} \right) = \sin^{-1} \frac{56}{65} = \mathbf{RHS}$$Hence, Proved.
Question 7. $\tan^{-1} \frac{63}{16} = \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}$
Solution:
We convert both terms on the RHS to $\tan^{-1}$ and then sum them.
- Convert $\sin^{-1} \frac{5}{13}$ to $\tan^{-1}$:If $\sin \alpha = 5/13$, adjacent side $= 12$.$\sin^{-1} \frac{5}{13} = \tan^{-1} \frac{5}{12}$.
- Convert $\cos^{-1} \frac{3}{5}$ to $\tan^{-1}$:If $\cos \beta = 3/5$, opposite side $= 4$.$\cos^{-1} \frac{3}{5} = \tan^{-1} \frac{4}{3}$.
- Apply $\tan^{-1} x + \tan^{-1} y$ identity to RHS:$$RHS = \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{4}{3}$$$$RHS = \tan^{-1} \left( \frac{\frac{5}{12} + \frac{4}{3}}{1 – \frac{5}{12} \cdot \frac{4}{3}} \right) = \tan^{-1} \left( \frac{\frac{5 + 16}{12}}{1 – \frac{20}{36}} \right)$$$$RHS = \tan^{-1} \left( \frac{21/12}{\frac{36 – 20}{36}} \right) = \tan^{-1} \left( \frac{21/12}{16/36} \right)$$$$RHS = \tan^{-1} \left( \frac{21}{12} \cdot \frac{36}{16} \right) = \tan^{-1} \left( \frac{21 \cdot 3}{16} \right) = \tan^{-1} \frac{63}{16} = \mathbf{LHS}$$Hence, Proved.
Prove that:
Question 8. $\tan^{-1} \frac{1}{2} \sqrt{\frac{1-x}{1+x}} = \frac{1}{2} \cos^{-1} x$, $x \in [0, 1]$
Solution:
Use the trigonometric substitution $x = \cos 2\theta$. Then $2\theta = \cos^{-1} x$, and $\theta = \frac{1}{2} \cos^{-1} x$.
LHS: $\tan^{-1} \sqrt{\frac{1-x}{1+x}}$
Substitute $x = \cos 2\theta$:
$$LHS = \tan^{-1} \sqrt{\frac{1 – \cos 2\theta}{1 + \cos 2\theta}}$$
Use the half-angle formulas: $1 – \cos 2\theta = 2 \sin^2 \theta$ and $1 + \cos 2\theta = 2 \cos^2 \theta$.
$$LHS = \tan^{-1} \sqrt{\frac{2 \sin^2 \theta}{2 \cos^2 \theta}} = \tan^{-1} \sqrt{\tan^2 \theta}$$
Since $x \in [0, 1]$, $2\theta \in [0, \pi/2]$, so $\theta \in [0, \pi/4]$, which means $\tan \theta > 0$.
$$LHS = \tan^{-1} (\tan \theta) = \theta$$
Substitute $\theta = \frac{1}{2} \cos^{-1} x$:
$$LHS = \frac{1}{2} \cos^{-1} x = \mathbf{RHS}$$
Hence, Proved.
Question 9. $\cot^{-1} \left(\frac{\sqrt{1 + \sin x} + \sqrt{1 – \sin x}}{\sqrt{1 + \sin x} – \sqrt{1 – \sin x}}\right) = \frac{x}{2}$, $x \in \left(0, \frac{\pi}{4}\right)$
Solution:
We simplify the terms under the square root. Use $\sin^2(x/2) + \cos^2(x/2) = 1$ and $2 \sin(x/2) \cos(x/2) = \sin x$.
$$1 + \sin x = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = \left(\cos \frac{x}{2} + \sin \frac{x}{2}\right)^2$$
$$1 – \sin x = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} – 2 \sin \frac{x}{2} \cos \frac{x}{2} = \left(\cos \frac{x}{2} – \sin \frac{x}{2}\right)^2$$
Since $x \in (0, \pi/4)$, $x/2 \in (0, \pi/8)$. In this range, $\cos(x/2) > \sin(x/2) > 0$.
$$\sqrt{1 + \sin x} = \cos \frac{x}{2} + \sin \frac{x}{2}$$
$$\sqrt{1 – \sin x} = \cos \frac{x}{2} – \sin \frac{x}{2}$$
Substitute these into the LHS expression:
$$LHS = \cot^{-1} \left(\frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) + (\cos \frac{x}{2} – \sin \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2}) – (\cos \frac{x}{2} – \sin \frac{x}{2})}\right)$$
$$LHS = \cot^{-1} \left(\frac{2 \cos \frac{x}{2}}{2 \sin \frac{x}{2}}\right) = \cot^{-1} \left(\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}\right)$$
$$LHS = \cot^{-1} \left(\cot \frac{x}{2}\right) = \frac{x}{2} = \mathbf{RHS}$$
Hence, Proved.
Question 10. $\tan^{-1} \left(\frac{\sqrt{1+x} – \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}\right) = \frac{\pi}{4} – \frac{1}{2} \cos^{-1} x$, $-\frac{1}{\sqrt{2}} \le x \le 1$
Solution:
Use the hint: Substitute $x = \cos 2\theta$. Then $2\theta = \cos^{-1} x$, and $\theta = \frac{1}{2} \cos^{-1} x$.
- Substitute using half-angle identities:$$1 + x = 1 + \cos 2\theta = 2 \cos^2 \theta$$$$1 – x = 1 – \cos 2\theta = 2 \sin^2 \theta$$$$\sqrt{1+x} = \sqrt{2} \cos \theta$$$$\sqrt{1-x} = \sqrt{2} \sin \theta$$
- Simplify the LHS:$$LHS = \tan^{-1} \left(\frac{\sqrt{2} \cos \theta – \sqrt{2} \sin \theta}{\sqrt{2} \cos \theta + \sqrt{2} \sin \theta}\right)$$Cancel $\sqrt{2}$:$$LHS = \tan^{-1} \left(\frac{\cos \theta – \sin \theta}{\cos \theta + \sin \theta}\right)$$Divide numerator and denominator by $\cos \theta$ (as in Q8, Ex 2.2):$$LHS = \tan^{-1} \left(\frac{1 – \tan \theta}{1 + \tan \theta}\right)$$$$LHS = \tan^{-1} \left(\tan \left(\frac{\pi}{4} – \theta\right)\right)$$$$LHS = \frac{\pi}{4} – \theta$$
- Substitute back $\theta$:$$LHS = \frac{\pi}{4} – \frac{1}{2} \cos^{-1} x = \mathbf{RHS}$$Hence, Proved.
Solve the following equations:
Question 11. $2\tan^{-1} (\cos x) = \tan^{-1} (2 \csc x)$
Solution:
Use the identity $2 \tan^{-1} A = \tan^{-1} \left(\frac{2A}{1-A^2}\right)$, where $A = \cos x$.
$$LHS = \tan^{-1} \left(\frac{2 \cos x}{1 – \cos^2 x}\right)$$
Since $1 – \cos^2 x = \sin^2 x$:
$$LHS = \tan^{-1} \left(\frac{2 \cos x}{\sin^2 x}\right)$$
Now set $LHS = RHS$:
$$\tan^{-1} \left(\frac{2 \cos x}{\sin^2 x}\right) = \tan^{-1} (2 \csc x)$$
Equating the arguments:
$$\frac{2 \cos x}{\sin^2 x} = 2 \csc x$$
Since $\csc x = 1/\sin x$:
$$\frac{2 \cos x}{\sin^2 x} = \frac{2}{\sin x}$$
Assuming $\sin x \neq 0$, multiply both sides by $\sin^2 x / 2$:
$$\cos x = \sin x$$
Divide by $\cos x$:
$$\tan x = 1$$
The principal value for which $\tan x = 1$ is $\pi/4$.
$$x = \frac{\pi}{4}$$
Question 12. $\tan^{-1} \frac{1-x}{1+x} = \frac{1}{2} \tan^{-1} x$, $(x > 0)$
Solution:
Use the identity $\tan^{-1} A – \tan^{-1} B = \tan^{-1} \left(\frac{A-B}{1+AB}\right)$.
We can rewrite $\tan^{-1} \frac{1-x}{1+x}$ as $\tan^{-1} \frac{1 – x}{1 + 1 \cdot x}$.
Since $\tan^{-1} 1 = \pi/4$, we have:
$$\tan^{-1} \left(\frac{1-x}{1+x}\right) = \tan^{-1} 1 – \tan^{-1} x = \frac{\pi}{4} – \tan^{-1} x$$
Substitute this back into the equation:
$$\frac{\pi}{4} – \tan^{-1} x = \frac{1}{2} \tan^{-1} x$$
Combine $\tan^{-1} x$ terms:
$$\frac{\pi}{4} = \tan^{-1} x + \frac{1}{2} \tan^{-1} x = \frac{3}{2} \tan^{-1} x$$
Solve for $\tan^{-1} x$:
$$\tan^{-1} x = \frac{\pi}{4} \cdot \frac{2}{3} = \frac{\pi}{6}$$
Take $\tan$ on both sides:
$$x = \tan \left(\frac{\pi}{6}\right)$$
$$x = \frac{1}{\sqrt{3}}$$
Multiple Choice Questions:
Question 13. $\sin (\tan^{-1} x)$, $|x| < 1$ is equal to
(A) $\frac{x}{\sqrt{1-x^2}}$ (B) $\frac{1}{\sqrt{1-x^2}}$ (C) $\frac{1}{\sqrt{1+x^2}}$ (D) $\frac{x}{\sqrt{1+x^2}}$
Solution:
We need to evaluate $\sin (\tan^{-1} x)$.
Let $\tan^{-1} x = \theta$. Then $\tan \theta = x = x/1$.
Construct a right triangle with opposite side $x$ and adjacent side $1$.
Hypotenuse $= \sqrt{x^2 + 1^2} = \sqrt{1+x^2}$.
Now, $\sin \theta$ is opposite/hypotenuse:
$$\sin (\tan^{-1} x) = \sin \theta = \frac{x}{\sqrt{1+x^2}}$$
Correct Answer: (D) $\frac{x}{\sqrt{1+x^2}}$
Question 14. $\sin^{-1} (1 – x) – 2 \sin^{-1} x = \frac{\pi}{2}$, then $x$ is equal to
(A) $0, 1/2$ (B) $1, 1/2$ (C) $0$ (D) $1/2$
Solution:
Given: $\sin^{-1} (1 – x) – 2 \sin^{-1} x = \frac{\pi}{2}$.
Rearrange the equation:
$$\sin^{-1} (1 – x) = \frac{\pi}{2} + 2 \sin^{-1} x$$
Take sine on both sides:
$$1 – x = \sin \left(\frac{\pi}{2} + 2 \sin^{-1} x\right)$$
Using the identity $\sin(\pi/2 + \theta) = \cos \theta$, where $\theta = 2 \sin^{-1} x$:
$$1 – x = \cos (2 \sin^{-1} x)$$
Let $\sin^{-1} x = y$. Then $x = \sin y$.
$$1 – x = \cos (2y)$$
Use the identity $\cos 2y = 1 – 2 \sin^2 y$:
$$1 – x = 1 – 2 \sin^2 y$$
Substitute $\sin y = x$:
$$1 – x = 1 – 2x^2$$
$$x = 2x^2$$
$$2x^2 – x = 0$$
Factor out $x$:
$$x(2x – 1) = 0$$
This gives two possible solutions: $x = 0$ or $x = 1/2$.
Now, we must verify these solutions in the original equation:
- Check $x = 0$:$$LHS = \sin^{-1} (1 – 0) – 2 \sin^{-1} (0) = \sin^{-1} (1) – 0 = \frac{\pi}{2}$$$$RHS = \frac{\pi}{2}$$$x=0$ is a valid solution.
- Check $x = 1/2$:$$LHS = \sin^{-1} (1 – 1/2) – 2 \sin^{-1} (1/2)$$$$LHS = \sin^{-1} (1/2) – 2 (\pi/6) = \frac{\pi}{6} – \frac{\pi}{3} = -\frac{\pi}{6}$$$$RHS = \frac{\pi}{2}$$Since $-\pi/6 \neq \pi/2$, $x = 1/2$ is an extraneous solution.
The only valid solution is $x = 0$.
Correct Answer: (C) $0$