Class 12 Mathematics Chapter 2 Question Answer | প্ৰতিলোম ত্ৰিকোণমিতীয় ফলন | English Medium | ASSEB

Inverse Trigonometric Functions — Questions and Answers

Welcome to HSLC Guru. This lesson gives complete, step-by-step solutions to every question of ASSEB Class 12 Mathematics Chapter 2 — Inverse Trigonometric Functions, covering Exercise 2.1, Exercise 2.2 and the Miscellaneous Exercise on Chapter 2, with each result worked out in full and every MCQ justified.

Summary

The six trigonometric functions are not one-one over their natural domains, so their inverses do not exist directly. By restricting the domain we make each function one-one and onto, and the inverse defined on that restricted piece is the principal value branch. The value of an inverse trigonometric function that lies inside its principal value branch is called the principal value.

The domains and principal value branches (ranges) of the inverse trigonometric functions are:

Function	Domain	Range (principal value branch)
$y = \sin^{-1} x$	$[-1, 1]$	$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$y = \cos^{-1} x$	$[-1, 1]$	$[0, \pi]$
$y = \operatorname{cosec}^{-1} x$	$\mathbb{R} – (-1, 1)$	$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] – \{0\}$
$y = \sec^{-1} x$	$\mathbb{R} – (-1, 1)$	$[0, \pi] – \left\{\frac{\pi}{2}\right\}$
$y = \tan^{-1} x$	$\mathbb{R}$	$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
$y = \cot^{-1} x$	$\mathbb{R}$	$(0, \pi)$

Two useful facts follow from the definition: $\sin(\sin^{-1} x) = x$ for $x \in [-1, 1]$, and $\sin^{-1}(\sin x) = x$ only when $x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$; similar relations hold for the other functions. Note that $\sin^{-1} x$ is not the same as $(\sin x)^{-1} = \frac{1}{\sin x}$. The graphs of an inverse trigonometric function and its parent function are mirror images in the line $y = x$; the principal value branches are shown below.

Summary: ASSEB Class 12 Mathematics Chapter 2, Inverse Trigonometric Functions, explains how restricting domains gives the principal value branches of sin⁻¹, cos⁻¹, tan⁻¹, cosec⁻¹, sec⁻¹ and cot⁻¹, and solves every question of Exercise 2.1, Exercise 2.2 and the Miscellaneous Exercise with full working, principal-value evaluations, simplification of inverse-trig expressions and proofs of standard identities.

Textbook Questions and Answers

Exercise 2.1

Find the principal values of the following (Questions 1–10).

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

Answer: Let $\sin^{-1}\left(-\frac{1}{2}\right) = y$, so $\sin y = -\frac{1}{2}$ where the principal value branch requires $y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Since $\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$ and $-\frac{\pi}{6}$ lies in this range, the principal value is $-\frac{\pi}{6}$.

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

Answer: Let $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = y$, so $\cos y = \frac{\sqrt{3}}{2}$ with $y \in [0, \pi]$. As $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $\frac{\pi}{6} \in [0, \pi]$, the principal value is $\frac{\pi}{6}$.

3. $\operatorname{cosec}^{-1}(2)$

Answer: Let $\operatorname{cosec}^{-1}(2) = y$, so $\operatorname{cosec} y = 2$, i.e. $\sin y = \frac{1}{2}$, with $y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] – \{0\}$. Since $\operatorname{cosec}\frac{\pi}{6} = 2$, the principal value is $\frac{\pi}{6}$.

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

Answer: Let $\tan^{-1}(-\sqrt{3}) = y$, so $\tan y = -\sqrt{3}$ with $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. As $\tan\left(-\frac{\pi}{3}\right) = -\sqrt{3}$, the principal value is $-\frac{\pi}{3}$.

5. $\cos^{-1}\left(-\frac{1}{2}\right)$

Answer: Let $\cos^{-1}\left(-\frac{1}{2}\right) = y$, so $\cos y = -\frac{1}{2}$ with $y \in [0, \pi]$. Since $\cos\frac{2\pi}{3} = -\frac{1}{2}$ and $\frac{2\pi}{3} \in [0, \pi]$, the principal value is $\frac{2\pi}{3}$.

6. $\tan^{-1}(-1)$

Answer: Let $\tan^{-1}(-1) = y$, so $\tan y = -1$ with $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. As $\tan\left(-\frac{\pi}{4}\right) = -1$, the principal value is $-\frac{\pi}{4}$.

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

Answer: Let $\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = y$, so $\sec y = \frac{2}{\sqrt{3}}$, i.e. $\cos y = \frac{\sqrt{3}}{2}$, with $y \in [0, \pi] – \left\{\frac{\pi}{2}\right\}$. Since $\sec\frac{\pi}{6} = \frac{2}{\sqrt{3}}$, the principal value is $\frac{\pi}{6}$.

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

Answer: Let $\cot^{-1}(\sqrt{3}) = y$, so $\cot y = \sqrt{3}$ with $y \in (0, \pi)$. As $\cot\frac{\pi}{6} = \sqrt{3}$, the principal value is $\frac{\pi}{6}$.

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

Answer: Let $\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = y$, so $\cos y = -\frac{1}{\sqrt{2}}$ with $y \in [0, \pi]$. Since $\cos\frac{3\pi}{4} = -\frac{1}{\sqrt{2}}$ and $\frac{3\pi}{4} \in [0, \pi]$, the principal value is $\frac{3\pi}{4}$.

10. $\operatorname{cosec}^{-1}(-\sqrt{2})$

Answer: Let $\operatorname{cosec}^{-1}(-\sqrt{2}) = y$, so $\operatorname{cosec} y = -\sqrt{2}$, i.e. $\sin y = -\frac{1}{\sqrt{2}}$, with $y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] – \{0\}$. Since $\operatorname{cosec}\left(-\frac{\pi}{4}\right) = -\sqrt{2}$, the principal value is $-\frac{\pi}{4}$.

Find the values of the following (Questions 11–12).

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

Answer: Evaluating each principal value: $\tan^{-1}(1) = \frac{\pi}{4}$, $\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$ and $\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$. Adding,

$$\frac{\pi}{4} + \frac{2\pi}{3} – \frac{\pi}{6} = \frac{3\pi + 8\pi – 2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}$$

Hence the value is $\frac{3\pi}{4}$.

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

Answer: Here $\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$ and $\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$. Therefore

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

Hence the value is $\frac{2\pi}{3}$.

13. If $\sin^{-1} x = y$, then

(A) $0 \le y \le \pi$ (B) $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$ (C) $0 < y < \pi$ (D) $-\frac{\pi}{2} < y < \frac{\pi}{2}$

Answer: (B). The principal value branch of $\sin^{-1}$ is the closed interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, and it is a closed interval because $\sin^{-1}(\pm 1) = \pm\frac{\pi}{2}$ are attained. Hence $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$. Option (D) is wrong because the end points are included; options (A) and (C) give the range of $\cos^{-1}$/$\cot^{-1}$, not $\sin^{-1}$.

14. $\tan^{-1}(\sqrt{3}) – \sec^{-1}(-2)$ is equal to

(A) $\pi$ (B) $-\frac{\pi}{3}$ (C) $\frac{\pi}{3}$ (D) $\frac{2\pi}{3}$

Answer: (B). We have $\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$. For $\sec^{-1}(-2) = y$, $\sec y = -2$ means $\cos y = -\frac{1}{2}$ with $y \in [0, \pi] – \left\{\frac{\pi}{2}\right\}$, giving $y = \frac{2\pi}{3}$. Therefore

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

Exercise 2.2

Prove the following (Questions 1–2).

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

Answer: Put $x = \sin\theta$, so that $\theta = \sin^{-1} x$. For $x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$ we have $\theta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right]$, hence $3\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Using the triple-angle identity,

$$\sin 3\theta = 3\sin\theta – 4\sin^3\theta = 3x – 4x^3$$

Since $3\theta$ lies in the principal value branch $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, we may write $\sin^{-1}(\sin 3\theta) = 3\theta$. Therefore $\sin^{-1}(3x – 4x^3) = 3\theta = 3\sin^{-1} x$, which proves the result.

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

Answer: Put $x = \cos\theta$, so $\theta = \cos^{-1} x$. For $x \in \left[\frac{1}{2}, 1\right]$ we have $\theta \in \left[0, \frac{\pi}{3}\right]$, hence $3\theta \in [0, \pi]$. By the triple-angle identity,

$$\cos 3\theta = 4\cos^3\theta – 3\cos\theta = 4x^3 – 3x$$

Since $3\theta \in [0, \pi]$, the principal value branch of $\cos^{-1}$, we get $\cos^{-1}(4x^3 – 3x) = 3\theta = 3\cos^{-1} x$, as required.

Write the following functions in the simplest form (Questions 3–7).

3. $\tan^{-1}\dfrac{\sqrt{1 + x^2} – 1}{x}$, $x \ne 0$.

Answer: Put $x = \tan\theta$, so $\theta = \tan^{-1} x$. Then $\sqrt{1 + x^2} = \sec\theta$, and

$$\frac{\sqrt{1 + x^2} – 1}{x} = \frac{\sec\theta – 1}{\tan\theta} = \frac{1 – \cos\theta}{\sin\theta} = \frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} = \tan\frac{\theta}{2}$$

Therefore $\tan^{-1}\dfrac{\sqrt{1 + x^2} – 1}{x} = \tan^{-1}\left(\tan\dfrac{\theta}{2}\right) = \dfrac{\theta}{2} = \dfrac{1}{2}\tan^{-1} x$.

4. $\tan^{-1}\sqrt{\dfrac{1 – \cos x}{1 + \cos x}}$, $0 < x < \pi$.

Answer: Using $1 – \cos x = 2\sin^2\frac{x}{2}$ and $1 + \cos x = 2\cos^2\frac{x}{2}$,

$$\sqrt{\frac{1 – \cos x}{1 + \cos x}} = \sqrt{\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}} = \left|\tan\frac{x}{2}\right| = \tan\frac{x}{2}$$

the last step holding because $0 < x < \pi$ gives $\frac{x}{2} \in \left(0, \frac{\pi}{2}\right)$, where $\tan\frac{x}{2} > 0$. Hence the expression equals $\tan^{-1}\left(\tan\frac{x}{2}\right) = \frac{x}{2}$.

5. $\tan^{-1}\left(\dfrac{\cos x – \sin x}{\cos x + \sin x}\right)$, $-\frac{\pi}{4} < x < \frac{3\pi}{4}$.

Answer: Divide numerator and denominator by $\cos x$:

$$\frac{\cos x – \sin x}{\cos x + \sin x} = \frac{1 – \tan x}{1 + \tan x} = \tan\left(\frac{\pi}{4} – x\right)$$

For $-\frac{\pi}{4} < x < \frac{3\pi}{4}$, the angle $\frac{\pi}{4} – x$ lies in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, so $\tan^{-1}\left[\tan\left(\frac{\pi}{4} – x\right)\right] = \frac{\pi}{4} – x$. Thus the simplest form is $\frac{\pi}{4} – x$.

6. $\tan^{-1}\dfrac{x}{\sqrt{a^2 – x^2}}$, $|x| < a$.

Answer: Put $x = a\sin\theta$, so $\theta = \sin^{-1}\frac{x}{a}$ with $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then $\sqrt{a^2 – x^2} = a\cos\theta$, and

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

Hence the expression equals $\tan^{-1}(\tan\theta) = \theta = \sin^{-1}\dfrac{x}{a}$.

7. $\tan^{-1}\left(\dfrac{3a^2 x – x^3}{a^3 – 3ax^2}\right)$, $a > 0$; $-\frac{a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$.

Answer: Put $x = a\tan\theta$, so $\theta = \tan^{-1}\frac{x}{a}$. Then

$$\frac{3a^2 x – x^3}{a^3 – 3ax^2} = \frac{a^3(3\tan\theta – \tan^3\theta)}{a^3(1 – 3\tan^2\theta)} = \frac{3\tan\theta – \tan^3\theta}{1 – 3\tan^2\theta} = \tan 3\theta$$

The condition $-\frac{a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$ gives $-\frac{1}{\sqrt{3}} < \tan\theta < \frac{1}{\sqrt{3}}$, i.e. $-\frac{\pi}{6} < \theta < \frac{\pi}{6}$, so $3\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Therefore the expression equals $\tan^{-1}(\tan 3\theta) = 3\theta = 3\tan^{-1}\dfrac{x}{a}$.

Find the values of each of the following (Questions 8–9).

8. $\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2}\right)\right]$

Answer: Since $\sin^{-1}\frac{1}{2} = \frac{\pi}{6}$, we get $2\sin^{-1}\frac{1}{2} = \frac{\pi}{3}$. Then $\cos\frac{\pi}{3} = \frac{1}{2}$, so the inner value is $2\cos\frac{\pi}{3} = 2 \cdot \frac{1}{2} = 1$. Finally $\tan^{-1}(1) = \frac{\pi}{4}$. Hence the value is $\frac{\pi}{4}$.

9. $\tan\dfrac{1}{2}\left[\sin^{-1}\dfrac{2x}{1 + x^2} + \cos^{-1}\dfrac{1 – y^2}{1 + y^2}\right]$, $|x| < 1$, $y > 0$ and $xy < 1$.

Answer: Using the standard identities $\sin^{-1}\dfrac{2x}{1 + x^2} = 2\tan^{-1} x$ (for $|x| \le 1$) and $\cos^{-1}\dfrac{1 – y^2}{1 + y^2} = 2\tan^{-1} y$ (for $y \ge 0$), the bracket becomes $2\tan^{-1} x + 2\tan^{-1} y$. Therefore

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

Hence the value is $\dfrac{x + y}{1 – xy}$.

Find the values of each of the following (Questions 10–12).

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

Answer: Here $\frac{2\pi}{3} \notin \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so we cannot cancel directly. Since $\sin\frac{2\pi}{3} = \sin\left(\pi – \frac{2\pi}{3}\right) = \sin\frac{\pi}{3}$ and $\frac{\pi}{3} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,

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

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

Answer: Since $\frac{3\pi}{4} \notin \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, use $\tan\frac{3\pi}{4} = \tan\left(\frac{3\pi}{4} – \pi\right) = \tan\left(-\frac{\pi}{4}\right)$, and $-\frac{\pi}{4}$ lies in the principal branch. Hence

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

12. $\tan\left(\sin^{-1}\frac{3}{5} + \cot^{-1}\frac{3}{2}\right)$

Answer: Let $A = \sin^{-1}\frac{3}{5}$, so $\sin A = \frac{3}{5}$, $\cos A = \frac{4}{5}$ and $\tan A = \frac{3}{4}$. Let $B = \cot^{-1}\frac{3}{2}$, so $\cot B = \frac{3}{2}$ and $\tan B = \frac{2}{3}$. Then

$$\tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} = \frac{\frac{3}{4} + \frac{2}{3}}{1 – \frac{3}{4}\cdot\frac{2}{3}} = \frac{\frac{17}{12}}{\frac{1}{2}} = \frac{17}{6}$$

Hence the value is $\dfrac{17}{6}$.

13. $\cos^{-1}\left(\cos\frac{7\pi}{6}\right)$ is equal to

(A) $\frac{7\pi}{6}$ (B) $\frac{5\pi}{6}$ (C) $\frac{\pi}{3}$ (D) $\frac{\pi}{6}$

Answer: (B). As $\frac{7\pi}{6} \notin [0, \pi]$, write $\cos\frac{7\pi}{6} = \cos\left(2\pi – \frac{7\pi}{6}\right) = \cos\frac{5\pi}{6}$, and $\frac{5\pi}{6} \in [0, \pi]$. Therefore $\cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \frac{5\pi}{6}$.

14. $\sin\left(\frac{\pi}{3} – \sin^{-1}\left(-\frac{1}{2}\right)\right)$ is equal to

(A) $\frac{1}{2}$ (B) $\frac{1}{3}$ (C) $\frac{1}{4}$ (D) $1$

Answer: (D). Since $\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$,

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

15. $\tan^{-1}(\sqrt{3}) – \cot^{-1}(-\sqrt{3})$ is equal to

(A) $\pi$ (B) $-\frac{\pi}{2}$ (C) $0$ (D) $2\sqrt{3}$

Answer: (B). Here $\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$. For $\cot^{-1}(-\sqrt{3}) = y$, $\cot y = -\sqrt{3}$ with $y \in (0, \pi)$, which gives $y = \pi – \frac{\pi}{6} = \frac{5\pi}{6}$. Therefore

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

Miscellaneous Exercise on Chapter 2

Find the value of the following (Questions 1–2).

1. $\cos^{-1}\left(\cos\frac{13\pi}{6}\right)$

Answer: Since $\frac{13\pi}{6} = 2\pi + \frac{\pi}{6}$, we have $\cos\frac{13\pi}{6} = \cos\frac{\pi}{6}$, and $\frac{\pi}{6} \in [0, \pi]$. Hence

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

2. $\tan^{-1}\left(\tan\frac{7\pi}{6}\right)$

Answer: As $\frac{7\pi}{6} \notin \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, use $\tan\frac{7\pi}{6} = \tan\left(\frac{7\pi}{6} – \pi\right) = \tan\frac{\pi}{6}$, with $\frac{\pi}{6}$ in the principal branch. Hence

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

Prove that (Questions 3–7).

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

Answer: Let $\sin^{-1}\frac{3}{5} = \theta$, so $\sin\theta = \frac{3}{5}$, $\cos\theta = \frac{4}{5}$ and $\tan\theta = \frac{3}{4}$, with $\theta \in \left(0, \frac{\pi}{2}\right)$. Then

$$\tan 2\theta = \frac{2\tan\theta}{1 – \tan^2\theta} = \frac{2\cdot\frac{3}{4}}{1 – \frac{9}{16}} = \frac{\frac{3}{2}}{\frac{7}{16}} = \frac{24}{7}$$

Now $2\theta \in \left(0, \frac{\pi}{2}\right)$, which is inside the principal branch of $\tan^{-1}$, so $2\theta = \tan^{-1}\frac{24}{7}$. Therefore $2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}$.

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

Answer: With right triangles, $\sin^{-1}\frac{8}{17} = \tan^{-1}\frac{8}{15}$ (since $\sqrt{17^2 – 8^2} = 15$) and $\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{3}{4}$. As $\frac{8}{15}\cdot\frac{3}{4} = \frac{2}{5} < 1$,

$$\tan^{-1}\frac{8}{15} + \tan^{-1}\frac{3}{4} = \tan^{-1}\frac{\frac{8}{15} + \frac{3}{4}}{1 – \frac{8}{15}\cdot\frac{3}{4}} = \tan^{-1}\frac{\frac{77}{60}}{\frac{36}{60}} = \tan^{-1}\frac{77}{36}$$

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

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

Answer: Let $A = \cos^{-1}\frac{4}{5}$ and $B = \cos^{-1}\frac{12}{13}$, so $\cos A = \frac{4}{5}$, $\sin A = \frac{3}{5}$, $\cos B = \frac{12}{13}$, $\sin B = \frac{5}{13}$. Then

$$\cos(A + B) = \cos A \cos B – \sin A \sin B = \frac{4}{5}\cdot\frac{12}{13} – \frac{3}{5}\cdot\frac{5}{13} = \frac{48 – 15}{65} = \frac{33}{65}$$

Since $A, B \in \left(0, \frac{\pi}{2}\right)$ their sum $A + B \in (0, \pi)$, so $A + B = \cos^{-1}\frac{33}{65}$, proving the result.

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

Answer: Let $A = \cos^{-1}\frac{12}{13}$ and $B = \sin^{-1}\frac{3}{5}$, so $\sin A = \frac{5}{13}$, $\cos A = \frac{12}{13}$, $\sin B = \frac{3}{5}$, $\cos B = \frac{4}{5}$. Then

$$\sin(A + B) = \sin A \cos B + \cos A \sin B = \frac{5}{13}\cdot\frac{4}{5} + \frac{12}{13}\cdot\frac{3}{5} = \frac{20 + 36}{65} = \frac{56}{65}$$

As $A + B$ lies in $\left(0, \frac{\pi}{2}\right) \subset \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, we get $A + B = \sin^{-1}\frac{56}{65}$, as required.

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

Answer: Convert the right side to $\tan^{-1}$ form: $\sin^{-1}\frac{5}{13} = \tan^{-1}\frac{5}{12}$ and $\cos^{-1}\frac{3}{5} = \tan^{-1}\frac{4}{3}$. Since $\frac{5}{12}\cdot\frac{4}{3} = \frac{5}{9} < 1$,

$$\tan^{-1}\frac{5}{12} + \tan^{-1}\frac{4}{3} = \tan^{-1}\frac{\frac{5}{12} + \frac{4}{3}}{1 – \frac{5}{12}\cdot\frac{4}{3}} = \tan^{-1}\frac{\frac{21}{12}}{\frac{4}{9}} = \tan^{-1}\frac{189}{48} = \tan^{-1}\frac{63}{16}$$

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

Prove that (Questions 8–10).

8. $\tan^{-1}\sqrt{x} = \frac{1}{2}\cos^{-1}\left(\frac{1 – x}{1 + x}\right)$, $x \in [0, 1]$.

Answer: Put $\sqrt{x} = \tan\theta$, so $\theta = \tan^{-1}\sqrt{x}$ and $x = \tan^2\theta$; for $x \in [0, 1]$, $\theta \in \left[0, \frac{\pi}{4}\right]$. Then

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

Since $2\theta \in \left[0, \frac{\pi}{2}\right] \subset [0, \pi]$, $\cos^{-1}\left(\frac{1 – x}{1 + x}\right) = \cos^{-1}(\cos 2\theta) = 2\theta$. Therefore $\frac{1}{2}\cos^{-1}\left(\frac{1 – x}{1 + x}\right) = \theta = \tan^{-1}\sqrt{x}$.

9. $\cot^{-1}\left(\dfrac{\sqrt{1 + \sin x} + \sqrt{1 – \sin x}}{\sqrt{1 + \sin x} – \sqrt{1 – \sin x}}\right) = \dfrac{x}{2}$, $x \in \left(0, \frac{\pi}{4}\right)$.

Answer: Using $1 + \sin x = \left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2$ and $1 – \sin x = \left(\cos\frac{x}{2} – \sin\frac{x}{2}\right)^2$; for $x \in \left(0, \frac{\pi}{4}\right)$ we have $\cos\frac{x}{2} > \sin\frac{x}{2} > 0$, so

$$\sqrt{1 + \sin x} = \cos\frac{x}{2} + \sin\frac{x}{2}, \qquad \sqrt{1 – \sin x} = \cos\frac{x}{2} – \sin\frac{x}{2}$$

Then the numerator is $2\cos\frac{x}{2}$ and the denominator is $2\sin\frac{x}{2}$, so their ratio is $\cot\frac{x}{2}$. Hence

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

since $\frac{x}{2} \in \left(0, \frac{\pi}{8}\right) \subset (0, \pi)$, which proves the result.

10. $\tan^{-1}\left(\dfrac{\sqrt{1 + x} – \sqrt{1 – x}}{\sqrt{1 + x} + \sqrt{1 – x}}\right) = \dfrac{\pi}{4} – \dfrac{1}{2}\cos^{-1} x$, $-\frac{1}{\sqrt{2}} \le x \le 1$. [Hint: Put $x = \cos 2\theta$.]

Answer: Put $x = \cos 2\theta$, so $2\theta = \cos^{-1} x$ and $\theta = \frac{1}{2}\cos^{-1} x$; for $-\frac{1}{\sqrt{2}} \le x \le 1$, $\theta \in \left[0, \frac{3\pi}{8}\right]$. Then $1 + x = 2\cos^2\theta$ and $1 – x = 2\sin^2\theta$, so $\sqrt{1 + x} = \sqrt{2}\cos\theta$ and $\sqrt{1 – x} = \sqrt{2}\sin\theta$ (both non-negative on this range). Thus

$$\frac{\sqrt{1 + x} – \sqrt{1 – x}}{\sqrt{1 + x} + \sqrt{1 – x}} = \frac{\cos\theta – \sin\theta}{\cos\theta + \sin\theta} = \frac{1 – \tan\theta}{1 + \tan\theta} = \tan\left(\frac{\pi}{4} – \theta\right)$$

Because $\frac{\pi}{4} – \theta \in \left[-\frac{\pi}{8}, \frac{\pi}{4}\right] \subset \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, the expression equals $\frac{\pi}{4} – \theta = \frac{\pi}{4} – \frac{1}{2}\cos^{-1} x$.

Solve the following equations (Questions 11–12).

11. $2\tan^{-1}(\cos x) = \tan^{-1}(2\operatorname{cosec} x)$

Answer: Apply $2\tan^{-1} t = \tan^{-1}\dfrac{2t}{1 – t^2}$ to the left side with $t = \cos x$:

$$\tan^{-1}\frac{2\cos x}{1 – \cos^2 x} = \tan^{-1}\frac{2\cos x}{\sin^2 x}$$

Equating with the right side $\tan^{-1}\dfrac{2}{\sin x}$ gives $\dfrac{2\cos x}{\sin^2 x} = \dfrac{2}{\sin x}$. Cancelling and cross-multiplying, $2\cos x \sin x = 2\sin^2 x$, so $\cos x = \sin x$, i.e. $\tan x = 1$. Hence $x = \frac{\pi}{4}$.

12. $\tan^{-1}\dfrac{1 – x}{1 + x} = \dfrac{1}{2}\tan^{-1} x$, $(x > 0)$

Answer: Since $\tan^{-1}\dfrac{1 – x}{1 + x} = \tan^{-1} 1 – \tan^{-1} x = \dfrac{\pi}{4} – \tan^{-1} x$ (valid for $x > -1$), the equation becomes

$$\frac{\pi}{4} – \tan^{-1} x = \frac{1}{2}\tan^{-1} x \implies \frac{\pi}{4} = \frac{3}{2}\tan^{-1} x \implies \tan^{-1} x = \frac{\pi}{6}$$

Therefore $x = \tan\frac{\pi}{6} = \dfrac{1}{\sqrt{3}}$, which is positive and hence acceptable.

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) $\frac{x}{\sqrt{1 + x^2}}$

Answer: (D). Let $\theta = \tan^{-1} x$, so $\tan\theta = x$ with $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Drawing a right triangle with opposite side $x$ and adjacent side $1$, the hypotenuse is $\sqrt{1 + x^2}$, so $\sin\theta = \dfrac{x}{\sqrt{1 + x^2}}$.

14. $\sin^{-1}(1 – x) – 2\sin^{-1} x = \frac{\pi}{2}$, then $x$ is equal to

(A) $0, \frac{1}{2}$ (B) $1, \frac{1}{2}$ (C) $0$ (D) $\frac{1}{2}$

Answer: (C). Rewrite as $\sin^{-1}(1 – x) = \frac{\pi}{2} + 2\sin^{-1} x$ and take sine of both sides. Using $\sin\left(\frac{\pi}{2} + 2\sin^{-1} x\right) = \cos(2\sin^{-1} x) = 1 – 2x^2$,

$$1 – x = 1 – 2x^2 \implies 2x^2 – x = 0 \implies x(2x – 1) = 0$$

giving $x = 0$ or $x = \frac{1}{2}$. Testing $x = \frac{1}{2}$: $\sin^{-1}\frac{1}{2} – 2\sin^{-1}\frac{1}{2} = \frac{\pi}{6} – \frac{\pi}{3} = -\frac{\pi}{6} \ne \frac{\pi}{2}$, so it is rejected. Testing $x = 0$: $\sin^{-1} 1 – 0 = \frac{\pi}{2}$, which holds. Hence $x = 0$.

Additional Questions and Answers

Multiple Choice Questions

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

(A) $\frac{\pi}{6}$ (B) $\frac{\pi}{3}$ (C) $\frac{\pi}{4}$ (D) $\frac{2\pi}{3}$

Answer: (B) $\frac{\pi}{3}$, because $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\frac{\pi}{3} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

2. The principal value branch (range) of $\cos^{-1} x$ is

(A) $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (B) $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (C) $[0, \pi]$ (D) $(0, \pi)$

Answer: (C) $[0, \pi]$.

3. $\cos^{-1}\left(\cos\frac{5\pi}{4}\right)$ equals

(A) $\frac{5\pi}{4}$ (B) $\frac{3\pi}{4}$ (C) $-\frac{5\pi}{4}$ (D) $\frac{\pi}{4}$

Answer: (B) $\frac{3\pi}{4}$. Since $\frac{5\pi}{4} \notin [0, \pi]$ and $\cos\frac{5\pi}{4} = \cos\frac{3\pi}{4} = -\frac{1}{\sqrt{2}}$ with $\frac{3\pi}{4} \in [0, \pi]$.

4. $\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3)$ equals

(A) $\frac{\pi}{2}$ (B) $\pi$ (C) $0$ (D) $\frac{\pi}{4}$

Answer: (B) $\pi$. Here $\tan^{-1}(2) + \tan^{-1}(3) = \pi + \tan^{-1}\frac{2 + 3}{1 – 6} = \pi – \frac{\pi}{4} = \frac{3\pi}{4}$; adding $\tan^{-1}(1) = \frac{\pi}{4}$ gives $\pi$.

5. $\sin^{-1}\left(\sin\frac{3\pi}{4}\right)$ equals

(A) $\frac{3\pi}{4}$ (B) $\frac{\pi}{4}$ (C) $-\frac{\pi}{4}$ (D) $\frac{\pi}{2}$

Answer: (B) $\frac{\pi}{4}$, since $\sin\frac{3\pi}{4} = \sin\frac{\pi}{4}$ and $\frac{\pi}{4} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

6. The domain of $\sec^{-1} x$ is

(A) $[-1, 1]$ (B) $\mathbb{R}$ (C) $\mathbb{R} – (-1, 1)$ (D) $(-1, 1)$

Answer: (C) $\mathbb{R} – (-1, 1)$, i.e. all $x$ with $|x| \ge 1$.

7. $\cos^{-1}\left(\frac{1}{2}\right) – 2\sin^{-1}\left(\frac{1}{2}\right)$ equals

(A) $0$ (B) $\frac{\pi}{3}$ (C) $\frac{2\pi}{3}$ (D) $\pi$

Answer: (A) $0$, since $\cos^{-1}\frac{1}{2} = \frac{\pi}{3}$ and $2\sin^{-1}\frac{1}{2} = 2\cdot\frac{\pi}{6} = \frac{\pi}{3}$.

8. If $\tan^{-1} x + \tan^{-1} y = \frac{\pi}{4}$ with $xy < 1$, then $x + y + xy$ equals

(A) $0$ (B) $1$ (C) $-1$ (D) $\frac{\pi}{4}$

Answer: (B) $1$. From $\tan^{-1}\frac{x + y}{1 – xy} = \frac{\pi}{4}$ we get $\frac{x + y}{1 – xy} = 1$, so $x + y = 1 – xy$, giving $x + y + xy = 1$.

Fill in the Blanks

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

2. For $x \in [-1, 1]$, $\sin^{-1} x + \cos^{-1} x = $ ______. Answer: $\frac{\pi}{2}$.

3. The range of the principal value branch of $\cot^{-1} x$ is ______. Answer: $(0, \pi)$.

4. The principal value of $\sec^{-1}(\sqrt{2})$ is ______. Answer: $\frac{\pi}{4}$.

5. For $x \in [-1, 1]$, $\cos^{-1}(-x) = $ ______. Answer: $\pi – \cos^{-1} x$.

True or False

1. $\sin^{-1} x$ is the same as $(\sin x)^{-1}$. Answer: False — $\sin^{-1} x$ is the inverse function, whereas $(\sin x)^{-1} = \frac{1}{\sin x}$.

2. $\tan^{-1}(\tan x) = x$ for every real number $x$. Answer: False — it holds only when $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

3. The principal value of $\cos^{-1}(-1)$ is $\pi$. Answer: True — $\cos\pi = -1$ and $\pi \in [0, \pi]$.

4. $\operatorname{cosec}^{-1} x$ is defined for every real number $x$. Answer: False — it is defined only for $|x| \ge 1$.

5. $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$ for all real $x$. Answer: True.

Short Answer Questions

1. Find the value of $\sin^{-1}\frac{1}{2} + \cos^{-1}\frac{1}{2}$.

Answer: $\sin^{-1}\frac{1}{2} = \frac{\pi}{6}$ and $\cos^{-1}\frac{1}{2} = \frac{\pi}{3}$, so the sum is $\frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$. (This also follows from the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$.)

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

Answer: As $\frac{2\pi}{3} \notin \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, use $\tan\frac{2\pi}{3} = \tan\left(\frac{2\pi}{3} – \pi\right) = \tan\left(-\frac{\pi}{3}\right)$. Since $-\frac{\pi}{3}$ is in the principal branch, $\tan^{-1}\left(\tan\frac{2\pi}{3}\right) = -\frac{\pi}{3}$.

3. Prove that $\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \frac{\pi}{4}$.

Answer: Since $\frac{1}{2}\cdot\frac{1}{3} = \frac{1}{6} < 1$, we may add directly:

$$\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \tan^{-1}\frac{\frac{1}{2} + \frac{1}{3}}{1 – \frac{1}{2}\cdot\frac{1}{3}} = \tan^{-1}\frac{\frac{5}{6}}{\frac{5}{6}} = \tan^{-1}(1) = \frac{\pi}{4}$$

4. Find the value of $\cos\left(\sin^{-1}\frac{3}{5} + \cos^{-1}\frac{12}{13}\right)$.

Answer: Let $A = \sin^{-1}\frac{3}{5}$ (so $\sin A = \frac{3}{5}$, $\cos A = \frac{4}{5}$) and $B = \cos^{-1}\frac{12}{13}$ (so $\cos B = \frac{12}{13}$, $\sin B = \frac{5}{13}$). Then