CBSE Class 12 Mathematics Inverse Trigonometric Functions VBQs Set A

BASIC CONCEPTS

Definition: If \( f : X \to Y \) is one-one onto (bijective) function, then there exists a unique function \( f^{-1} : Y \to X \) which assigns each element \( y \in Y \) to a unique element \( x \in X \) such that \( f(x) = y \) and is called inverse function of \( f \).
i.e., \( f^{-1}(y) = x \Leftrightarrow f(x) = y, x \in X \) and \( y \in Y \)

Principal value branches: Since trigonometric functions being periodic are in general not bijective (one-one onto) and thus for existence of inverse of trigonometric function we restrict their domain and co-domain to make it bijective. This restriction of domain and range gives principal value branch of inverse trigonometric function which are as follows:

• \( y = \sin^{-1} x \), Domain: \( [-1, 1] \), Range (Principal value branch): \( [-\frac{\pi}{2}, \frac{\pi}{2}] \)
• \( y = \cos^{-1} x \), Domain: \( [-1, 1] \), Range (Principal value branch): \( [0, \pi] \)
• \( y = \csc^{-1} x \), Domain: \( \mathbb{R} - (-1, 1) \), Range (Principal value branch): \( [-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\} \)
• \( y = \sec^{-1} x \), Domain: \( \mathbb{R} - (-1, 1) \), Range (Principal value branch): \( [0, \pi] - \{\frac{\pi}{2}\} \)
• \( y = \tan^{-1} x \), Domain: \( \mathbb{R} \), Range (Principal value branch): \( (-\frac{\pi}{2}, \frac{\pi}{2}) \)
• \( y = \cot^{-1} x \), Domain: \( \mathbb{R} \), Range (Principal value branch): \( (0, \pi) \)

The value of an inverse trigonometric function which lies in its principal value branch is called the principal value of that inverse trigonometric function.

Principal and general values:
(a) If \( \sin \theta = \sin \alpha \) then its principal value is \( \theta = \alpha, -\frac{\pi}{2} \le \alpha \le \frac{\pi}{2} \) and its general value is \( \theta = n\pi + (-1)^n \alpha, n \in \mathbb{Z} \)
(b) If \( \cos \theta = \cos \alpha \) then its principal value is \( \theta = \alpha, 0 < \alpha < \pi \) and its general value is \( \theta = 2n\pi \pm \alpha, n \in \mathbb{Z} \)
(c) If \( \tan \theta = \tan \alpha \) then its principal value is \( \theta = \alpha, -\frac{\pi}{2} < \alpha < \frac{\pi}{2} \) and its general value is \( \theta = n\pi + \alpha, n \in \mathbb{Z} \)

Properties of Inverse Trigonometric Functions

1. (i) \( \sin^{-1}(\sin \theta) = \theta \), for all \( \theta \in [-\pi/2, \pi/2] \)
(ii) \( \cos^{-1}(\cos \theta) = \theta \), for all \( \theta \in [0, \pi] \)
(iii) \( \tan^{-1}(\tan \theta) = \theta \), for all \( \theta \in (-\pi/2, \pi/2) \)
(iv) \( \csc^{-1}(\csc \theta) = \theta \), for all \( \theta \in [-\pi/2, \pi/2], \theta \neq 0 \)
(v) \( \sec^{-1}(\sec \theta) = \theta \), for all \( \theta \in [0, \pi], \theta \neq \pi/2 \)
(vi) \( \cot^{-1}(\cot \theta) = \theta \), for all \( \theta \in (0, \pi) \)

2. (i) \( \sin(\sin^{-1} x) = x \), for all \( x \in [-1, 1] \)
(ii) \( \cos(\cos^{-1} x) = x \), for all \( x \in [-1, 1] \)
(iii) \( \tan(\tan^{-1} x) = x \), for all \( x \in \mathbb{R} \)
(iv) \( \csc(\csc^{-1} x) = x \), for all \( x \in (-\infty, -1] \cup [1, \infty) \)
(v) \( \sec(\sec^{-1} x) = x \), for all \( x \in (-\infty, -1] \cup [1, \infty) \)
(vi) \( \cot(\cot^{-1} x) = x \), for all \( x \in \mathbb{R} \)

3. (i) \( \sin^{-1} \left(\frac{1}{x}\right) = \csc^{-1} x \), for all \( x \in (-\infty, -1] \cup [1, \infty) \)
(ii) \( \cos^{-1} \left(\frac{1}{x}\right) = \sec^{-1} x \), for all \( x \in (-\infty, -1] \cup [1, \infty) \)
(iii) \( \tan^{-1} \left(\frac{1}{x}\right) = \begin{cases} \cot^{-1} x, & \text{for all } x > 0 \\ -\pi + \cot^{-1} x, & \text{for all } x < 0 \end{cases} \)

4. (i) \( \sin^{-1} (-x) = -\sin^{-1} x \), for all \( x \in [-1, 1] \)
(ii) \( \cos^{-1} (-x) = \pi - \cos^{-1} x \), for all \( x \in [-1, 1] \)
(iii) \( \tan^{-1} (-x) = -\tan^{-1} x \), for all \( x \in \mathbb{R} \)
(iv) \( \csc^{-1} (-x) = -\csc^{-1} x \), for all \( x \in (-\infty, -1] \cup [1, \infty) \)
(v) \( \sec^{-1} (-x) = \pi - \sec^{-1} x \), for all \( x \in (-\infty, -1] \cup [1, \infty) \)
(vi) \( \cot^{-1} (-x) = \pi - \cot^{-1} x \), for all \( x \in \mathbb{R} \)

5. (i) \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \), for all \( x \in [-1, 1] \)
(ii) \( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \), for all \( x \in \mathbb{R} \)
(iii) \( \sec^{-1} x + \csc^{-1} x = \frac{\pi}{2} \), for all \( x \in (-\infty, -1] \cup [1, \infty) \)

6. (i) \( \sin^{-1} x + \sin^{-1} y = \begin{cases} \sin^{-1} \{x \sqrt{1-y^2} + y \sqrt{1-x^2}\}, & \text{if } -1 \le x, y \le 1 \text{ and } x^2 + y^2 \le 1 \\ \text{or} & \text{if } xy < 0 \text{ and } x^2 + y^2 > 1 \end{cases} \)
(ii) \( \sin^{-1} x - \sin^{-1} y = \begin{cases} \sin^{-1} \{x \sqrt{1-y^2} - y \sqrt{1-x^2}\}, & \text{if } -1 \le x, y \le 1 \text{ and } x^2 + y^2 \le 1 \\ \text{or} & \text{if } xy > 0 \text{ and } x^2 + y^2 > 1 \end{cases} \)

7. (i) \( \cos^{-1} x + \cos^{-1} y = \cos^{-1} \{xy - \sqrt{1-x^2}\sqrt{1-y^2}\} \), if \( -1 \le x, y \le 1 \) and \( x + y \ge 0 \)
(ii) \( \cos^{-1} x - \cos^{-1} y = \cos^{-1} \{xy + \sqrt{1-x^2}\sqrt{1-y^2}\} \), if \( -1 \le x, y \le 1 \) and \( x \le y \)

8. (i) \( \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x + y}{1 - xy} \right) \), if \( xy < 1 \)
(ii) \( \tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x - y}{1 + xy} \right) \), if \( xy > -1 \)

9. (i) \( 2 \sin^{-1} x = \sin^{-1} (2x\sqrt{1-x^2}) \), if \( -\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}} \)
(ii) \( 2 \cos^{-1} x = \cos^{-1} (2x^2 - 1) \), if \( 0 \le x \le 1 \)
(iii) \( 2 \tan^{-1} x = \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \), if \( -1 < x < 1 \)

10. (i) \( 3 \sin^{-1} x = \sin^{-1} (3x - 4x^3) \), if \( -\frac{1}{2} \le x \le \frac{1}{2} \)
(ii) \( 3 \cos^{-1} x = \cos^{-1} (4x^3 - 3x) \), if \( \frac{1}{2} \le x \le 1 \)
(iii) \( 3 \tan^{-1} x = \tan^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right) \), if \( -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}} \)

11. (i) \( 2 \tan^{-1} x = \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \), if \( -1 \le x \le 1 \)
(ii) \( 2 \tan^{-1} x = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \), if \( 0 \le x < \infty \)

12. (i) \( \sin^{-1} x = \cos^{-1} (\sqrt{1-x^2}) = \tan^{-1} \left( \frac{x}{\sqrt{1-x^2}} \right) = \cot^{-1} \left( \frac{\sqrt{1-x^2}}{x} \right) = \sec^{-1} \left( \frac{1}{\sqrt{1-x^2}} \right) = \csc^{-1} \left( \frac{1}{x} \right) \)
(ii) \( \cos^{-1} x = \sin^{-1} (\sqrt{1-x^2}) = \tan^{-1} \left( \frac{\sqrt{1-x^2}}{x} \right) = \cot^{-1} \left( \frac{x}{\sqrt{1-x^2}} \right) = \sec^{-1} \left( \frac{1}{x} \right) = \csc^{-1} \left( \frac{1}{\sqrt{1-x^2}} \right) \)
(iii) \( \tan^{-1} x = \sin^{-1} \left( \frac{x}{\sqrt{1+x^2}} \right) = \cos^{-1} \left( \frac{1}{\sqrt{1+x^2}} \right) = \cot^{-1} \left( \frac{1}{x} \right) = \sec^{-1} (\sqrt{1+x^2}) = \csc^{-1} \left( \frac{\sqrt{1+x^2}}{x} \right) \)

Important substitution to simplify trigonometrical expressions involving inverse trigonometrical functions.

• Expression: \( a^2 + x^2 \), Substitution: \( x = a \tan \theta \) or \( x = a \cot \theta \)
• Expression: \( a^2 - x^2 \), Substitution: \( x = a \sin \theta \) or \( x = a \cos \theta \)
• Expression: \( x^2 - a^2 \), Substitution: \( x = a \sec \theta \) or \( x = a \csc \theta \)
• Expression: \( \sqrt{\frac{a+x}{a-x}} \) or \( \sqrt{\frac{a-x}{a+x}} \), Substitution: \( x = a \cos 2\theta \)

Inverse Trigonometric Functions

Important Facts:

(i) If no branch of an inverse trigonometric function is mentioned, we mean the principal value branch of that function.

(ii) \( \sin^{-1} x \neq \frac{1}{\sin x} \) or \( (\sin x)^{-1} \) and same holds true for other trigonometric functions also.

(iii) If \( \sin^{-1} x = y \) then \( x \) and \( y \) are the elements of domain and range of principal value branch of \( \sin^{-1} \) respectively.
i.e., \( x \in [-1, 1] \) and \( y \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \)

Similar fact is also applicable for other inverse trigonometric functions.

Question. Find the principal values of \( \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) \).
Answer: Let \( \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) = \theta \) so that \( \theta \in [0, \pi] \) and \( \cos \theta = \frac{\sqrt{3}}{2} \).
\( \Rightarrow \cos \theta = \cos \frac{\pi}{6} \Rightarrow \theta = \frac{\pi}{6} \), i.e., \( \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{6} \)

Question. Find the principal values of \( \tan^{-1}(-\sqrt{3}) \).
Answer: Let \( \tan^{-1}(-\sqrt{3}) = \theta \) where \( \theta \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \)
\( \Rightarrow \tan \theta = -\sqrt{3} \Rightarrow \tan \theta = -\tan\left( \frac{\pi}{3} \right) \)
\( \Rightarrow \tan \theta = \tan \left( -\frac{\pi}{3} \right) \Rightarrow \theta = -\frac{\pi}{3} \)
\( \Rightarrow \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} \).

Question. Write the following function in the simplest form:
\( \tan^{-1} \left( \sqrt{\frac{1 - \cos x}{1 + \cos x}} \right), 0 < x < \pi \)
Answer: \( \tan^{-1} \left( \sqrt{\frac{1 - \cos x}{1 + \cos x}} \right) = \tan^{-1} \left( \sqrt{\frac{2 \sin^2 \left( \frac{x}{2} \right)}{2 \cos^2 \left( \frac{x}{2} \right)}} \right) \)
\( = \tan^{-1} \left( \left| \tan \left( \frac{x}{2} \right) \right| \right) = \tan^{-1} \left( \tan \frac{x}{2} \right) = \frac{x}{2} \)
[Note that \( 0 < x < \pi \Leftrightarrow 0 < \frac{x}{2} < \frac{\pi}{2} \)]

Question. Write the following function in the simplest form:
\( \tan^{-1} \left( \frac{\cos x - \sin x}{\cos x + \sin x} \right), -\frac{\pi}{4} < x < \frac{3\pi}{4} \)
Answer: \( \tan^{-1} \left( \frac{\cos x - \sin x}{\cos x + \sin x} \right) \)
Inside the bracket divide \( N^r \) and \( D^r \) by \( \cos x \)
\( = \tan^{-1} \left( \frac{1 - \tan x}{1 + \tan x} \right) = \tan^{-1} \left\{ \tan \left( \frac{\pi}{4} - x \right) \right\} = \frac{\pi}{4} - x \)
Note that \( -\frac{\pi}{4} < x < \frac{3\pi}{4} \Rightarrow \frac{\pi}{4} > -x > -\frac{3\pi}{4} \Rightarrow \frac{\pi}{2} > \frac{\pi}{4} - x > -\frac{\pi}{2} \)

Question. Find the value of the \( \tan \left\{ \frac{1}{2} \left[ \sin^{-1} \left( \frac{2x}{1 + x^2} \right) + \cos^{-1} \left( \frac{1 - y^2}{1 + y^2} \right) \right] \right\}, |x| < 1, y > 0 \text{ and } xy < 1. \)
Answer: \( \tan \left\{ \frac{1}{2} \left[ \sin^{-1} \left( \frac{2x}{1 + x^2} \right) + \cos^{-1} \left( \frac{1 - y^2}{1 + y^2} \right) \right] \right\} \)
\( = \tan \left\{ \frac{1}{2} (2 \tan^{-1} x + 2 \tan^{-1} y) \right\} \)
[\( \because \sin^{-1} \frac{2x}{1+x^2} = 2 \tan^{-1} x \) and \( \cos^{-1} \frac{1-y^2}{1+y^2} = 2 \tan^{-1} y \)]
\( = \tan \{ \tan^{-1} x + \tan^{-1} y \} = \tan \left\{ \tan^{-1} \left( \frac{x+y}{1-xy} \right) \right\} = \frac{x+y}{1-xy} \).

Question. Prove that: \( \tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{7} + \tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{8} = \frac{\pi}{4} \)
Answer: We know that \( \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right) \) for \( xy < 1 \), therefore,
LHS \( = \left( \tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{7} \right) + \left( \tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{8} \right) \)
\( = \tan^{-1} \left( \frac{\frac{1}{5} + \frac{1}{7}}{1 - \frac{1}{5} \times \frac{1}{7}} \right) + \tan^{-1} \left( \frac{\frac{1}{3} + \frac{1}{8}}{1 - \frac{1}{3} \times \frac{1}{8}} \right) \)
\( = \tan^{-1} \left( \frac{12}{34} \right) + \tan^{-1} \left( \frac{11}{23} \right) = \tan^{-1} \left( \frac{6}{17} \right) + \tan^{-1} \left( \frac{11}{23} \right) \)
\( = \tan^{-1} \left( \frac{\frac{6}{17} + \frac{11}{23}}{1 - \frac{6}{17} \times \frac{11}{23}} \right) = \tan^{-1} \left( \frac{6 \times 23 + 11 \times 17}{17 \times 23 - 6 \times 11} \right) \)
\( = \tan^{-1} \left( \frac{325}{325} \right) = \tan^{-1} 1 = \frac{\pi}{4} = \text{RHS} \)

Question. Prove that: \( \tan^{-1} \sqrt{x} = \frac{1}{2} \cos^{-1} \left( \frac{1 - x}{1 + x} \right), x \in [0, 1] \)
Answer: Let \( \tan^{-1} \sqrt{x} = \theta \) then \( \tan \theta = \sqrt{x} \) and \( 0 \le \theta < \frac{\pi}{2} \)
Now, \( \frac{1}{2} \cos^{-1} \left( \frac{1 - x}{1 + x} \right) = \frac{1}{2} \cos^{-1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) \)
[\( \because \cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \)]
\( = \frac{1}{2} \cos^{-1} (\cos 2\theta) = \frac{1}{2} (2\theta) = \theta = \tan^{-1} \sqrt{x} \)
[\( \because 0 \le \theta < \frac{\pi}{2} \Rightarrow 0 \le 2\theta < \pi \)]
Note that the result is valid for all \( x \ge 0 \). In particular, it is valid for all \( x \in [0, 1] \).

Question. Prove that: \( \cot^{-1} \left( \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} \right) = \frac{x}{2}, x \in \left( 0, \frac{\pi}{4} \right) \)
Answer: LHS \( = \cot^{-1} \left( \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} \right), x \in \left( 0, \frac{\pi}{4} \right) \)
\( = \cot^{-1} \left( \frac{\sqrt{(\cos x/2 + \sin x/2)^2} + \sqrt{(\cos x/2 - \sin x/2)^2}}{\sqrt{(\cos x/2 + \sin x/2)^2} - \sqrt{(\cos x/2 - \sin x/2)^2}} \right) \)
\( = \cot^{-1} \left\{ \frac{\left| \cos \frac{x}{2} + \sin \frac{x}{2} \right| + \left| \cos \frac{x}{2} - \sin \frac{x}{2} \right|}{\left| \cos \frac{x}{2} + \sin \frac{x}{2} \right| - \left| \cos \frac{x}{2} - \sin \frac{x}{2} \right|} \right\} \)
Given \( 0 < x < \frac{\pi}{4} \Rightarrow 0 < \frac{x}{2} < \frac{\pi}{8} \Rightarrow \cos \frac{x}{2} - \sin \frac{x}{2} > 0 \)
\( \Rightarrow \left| \cos \frac{x}{2} - \sin \frac{x}{2} \right| = \cos \frac{x}{2} - \sin \frac{x}{2} \)
\( = \cot^{-1} \left( \frac{(\cos x/2 + \sin x/2) + (\cos x/2 - \sin x/2)}{(\cos x/2 + \sin x/2) - (\cos x/2 - \sin x/2)} \right) \)
\( = \cot^{-1} \left( \frac{2 \cos x/2}{2 \sin x/2} \right) = \cot^{-1} (\cot x/2) = \frac{x}{2} = \text{RHS} \)

Question. Prove that: \( \tan^{-1} \left( \frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} \right) = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x, -\frac{1}{\sqrt{2}} \le x \le 1 \)
Answer: LHS \( = \tan^{-1} \left( \frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} \right) \)
\( = \tan^{-1} \left( \frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}} \times \frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} - \sqrt{1 - x}} \right) \) [Rationalize]
\( = \tan^{-1} \left( \frac{(1 + x) + (1 - x) - 2\sqrt{1 - x^2}}{1 + x - (1 - x)} \right) \)
\( = \tan^{-1} \left( \frac{2 - 2\sqrt{1 - x^2}}{2x} \right) = \tan^{-1} \left( \frac{1 - \sqrt{1 - x^2}}{x} \right) \)
Put \( x = \sin \theta \Rightarrow \theta = \sin^{-1} x \)
\( = \tan^{-1} \left( \frac{1 - \cos \theta}{\sin \theta} \right) = \tan^{-1} \left( \frac{2 \sin^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}} \right) \)
\( = \tan^{-1} \left( \tan \frac{\theta}{2} \right) = \frac{\theta}{2} = \frac{1}{2} \sin^{-1} x \)
\( = \frac{1}{2} \left( \frac{\pi}{2} - \cos^{-1} x \right) = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x = \text{RHS} \).
[\( \because \cos^{-1} x + \sin^{-1} x = \frac{\pi}{2} \)]
Given \( -\frac{1}{\sqrt{2}} \le x \le 1 \Rightarrow \sin\left(-\frac{\pi}{4}\right) \le \sin \theta \le \sin\left(\frac{\pi}{2}\right) \Rightarrow -\frac{\pi}{4} \le \theta \le \frac{\pi}{2} \Rightarrow -\frac{\pi}{8} \le \frac{\theta}{2} \le \frac{\pi}{4} \Rightarrow \frac{\theta}{2} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \)

Question. Prove that: \( \frac{9\pi}{8} - \frac{9}{4} \sin^{-1} \frac{1}{3} = \frac{9}{4} \sin^{-1} \frac{2\sqrt{2}}{3} \)
Answer: LHS \( = \frac{9\pi}{8} - \frac{9}{4} \sin^{-1} \frac{1}{3} = \frac{9}{4} \left( \frac{\pi}{2} - \sin^{-1} \frac{1}{3} \right) \)
\( = \frac{9}{4} \cos^{-1} \left( \frac{1}{3} \right) \) ...(i) [\( \because \frac{1}{3} \in [-1, 1] \)]
Let, \( \cos^{-1} \left( \frac{1}{3} \right) = \theta \Rightarrow \cos \theta = \frac{1}{3} \)
\( \Rightarrow \sin \theta = \sqrt{1 - \left( \frac{1}{3} \right)^2} \) [\( \because \theta \in [0, \pi] \Rightarrow \sin \theta \text{ is positive} \)]
\( \Rightarrow \sin \theta = \sqrt{\frac{8}{9}} \Rightarrow \sin \theta = \frac{2\sqrt{2}}{3} \)
\( \Rightarrow \theta = \sin^{-1} \frac{2\sqrt{2}}{3} \) [\( \because \frac{2\sqrt{2}}{3} \in [-1, 1] \)]
\( \Rightarrow \cos^{-1} \left( \frac{1}{3} \right) = \sin^{-1} \frac{2\sqrt{2}}{3} \)
\(\therefore\) From equation (i), we have \( \frac{9}{4} \sin^{-1} \frac{2\sqrt{2}}{3} = \text{RHS} \)

Question. Solve: \( \tan^{-1} \frac{1 - x}{1 + x} = \frac{1}{2} \tan^{-1} x, x > 0 \)
Answer: Given, \( \tan^{-1} \frac{1 - x}{1 + x} = \frac{1}{2} \tan^{-1} x \Rightarrow 2 \tan^{-1} \frac{1 - x}{1 + x} = \tan^{-1} x \)
\( \Rightarrow \tan^{-1} \frac{2\left(\frac{1 - x}{1 + x}\right)}{1 - \left(\frac{1 - x}{1 + x}\right)^2} = \tan^{-1} x \) \( \left[ \because x > 0 \Rightarrow -1 < \frac{1 - x}{1 + x} < 1 \right] \)
\( \Rightarrow \tan^{-1} \frac{2(1 - x^2)}{4x} = \tan^{-1} x \Rightarrow \frac{2(1 - x^2)}{4x} = x \)
\( \Rightarrow 1 - x^2 = 2x^2 \Rightarrow 1 = 3x^2 \)
i.e., \( x^2 = \frac{1}{3} \therefore x = \frac{1}{\sqrt{3}} \) [\( \because x > 0 \)]

Question. If \( \tan^{-1} \frac{x - 1}{x - 2} + \tan^{-1} \frac{x + 1}{x + 2} = \frac{\pi}{4} \), then find the value of x.
Answer: Given \( \tan^{-1} \frac{x - 1}{x - 2} + \tan^{-1} \frac{x + 1}{x + 2} = \frac{\pi}{4} \)
\( \Rightarrow \tan^{-1} \left[ \frac{\frac{x - 1}{x - 2} + \frac{x + 1}{x + 2}}{1 - \frac{x - 1}{x - 2} \times \frac{x + 1}{x + 2}} \right] = \frac{\pi}{4} \) \( \left[ \text{Using } \tan^{-1} x + \tan^{-1} y = \tan^{-1} \frac{x + y}{1 - xy} \right] \)
\( \Rightarrow \frac{(x - 1)(x + 2) + (x + 1)(x - 2)}{(x - 2)(x + 2) - (x - 1)(x + 1)} = \tan \frac{\pi}{4} \)
\( \Rightarrow \frac{x^2 + x - 2 + x^2 - x - 2}{x^2 - 4 - x^2 + 1} = 1 \Rightarrow \frac{2(x^2 - 2)}{-3} = 1 \Rightarrow 2x^2 - 4 = -3 \)
\( \Rightarrow 2x^2 = 1 \Rightarrow x^2 = \frac{1}{2} \)
\( \therefore x = \pm \frac{1}{\sqrt{2}} \)

Fill in the Blanks

Question. The value of \( \sin^{-1} \left( \sin \frac{3\pi}{5} \right) \) is ____________ .
Answer: We have, \( \sin^{-1} \left( \sin \frac{3\pi}{5} \right) = \sin^{-1} \left[ \sin \left( \pi - \frac{2\pi}{5} \right) \right] = \sin^{-1} \left( \sin \frac{2\pi}{5} \right) = \frac{2\pi}{5} \)

Question. The principal value of \( \tan^{-1}(-\sqrt{3}) \) is ____________ .
Answer: We have, \( \tan^{-1}(-\sqrt{3}) = \tan^{-1} [-\tan \frac{\pi}{3}] = \tan^{-1} [\tan (-\frac{\pi}{3})] = -\frac{\pi}{3} \)

Question. The value of \( \sin(\tan^{-1} 2 + \cot^{-1} 2) \) is ____________ .
Answer: We have, \( \sin(\tan^{-1} 2 + \cot^{-1} 2) = \sin \frac{\pi}{2} = 1 \) (since \( \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \))

Question. If \( \cos(\tan^{-1} x + \cot^{-1} \sqrt{3}) = 0 \), then value of \( x \) is ____________ .
Answer: We have, \( \cos(\tan^{-1} x + \cot^{-1} \sqrt{3}) = 0 = \cos \frac{\pi}{2} \)
\( \Rightarrow \tan^{-1} x + \cot^{-1} \sqrt{3} = \frac{\pi}{2} \)
\( \Rightarrow \cot^{-1} \sqrt{3} = \frac{\pi}{2} - \tan^{-1} x = \cot^{-1} x \)
\( \Rightarrow \cot^{-1} \sqrt{3} = \cot^{-1} x \Rightarrow x = \sqrt{3} \)

Question. The value of \( \sin^{-1} \left[ \cos \left\{ \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) \right\} \right] \) is ____________ .
Answer: We have, \( \sin^{-1} \left[ \cos \left\{ \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) \right\} \right] = \sin^{-1} [\cos \{ \sin^{-1} (\sin \frac{\pi}{3}) \}] = \sin^{-1} (\cos \frac{\pi}{3}) = \sin^{-1} (\frac{1}{2}) = \sin^{-1} (\sin \frac{\pi}{6}) = \frac{\pi}{6} \)

Question. The principal value of \( \cos^{-1} \left( -\frac{1}{2} \right) \) is ____________ . 
Answer: \( \cos^{-1} (-\frac{1}{2}) = \cos^{-1} [-\cos \frac{\pi}{3}] = \cos^{-1} [\cos (\pi - \frac{\pi}{3})] = \cos^{-1} (\cos \frac{2\pi}{3}) = \frac{2\pi}{3} \)

Very Short Answer Questions

Question. Find the value of \( \sin^{-1} \left[ \sin \left( -\frac{17\pi}{8} \right) \right] \). 
Answer: We have, \( \sin^{-1} \left[ \sin \left( -\frac{17\pi}{8} \right) \right] = \sin^{-1} [-\sin \frac{17\pi}{8}] = \sin^{-1} [-\sin (2\pi + \frac{\pi}{8})] = \sin^{-1} [-\sin \frac{\pi}{8}] = \sin^{-1} [\sin (-\frac{\pi}{8})] = -\frac{\pi}{8} \). (since \( -\frac{\pi}{8} \in [-\frac{\pi}{2}, \frac{\pi}{2}] \))

Question. Write the principal value of \( \tan^{-1} 1 + \cos^{-1} \left( -\frac{1}{2} \right) \). 
Answer: \( \tan^{-1} 1 + \cos^{-1} (-\frac{1}{2}) = \tan^{-1} (\tan \frac{\pi}{4}) + \cos^{-1} (\cos \frac{2\pi}{3}) = \frac{\pi}{4} + \frac{2\pi}{3} = \frac{3\pi + 8\pi}{12} = \frac{11\pi}{12} \). (since \( \frac{\pi}{4} \in (-\frac{\pi}{2}, \frac{\pi}{2}) \) and \( \frac{2\pi}{3} \in [0, \pi] \))

Question. Write the value of \( \tan \left( 2 \tan^{-1} \frac{1}{5} \right) \).
Answer: \( \tan(2 \tan^{-1} \frac{1}{5}) = \tan \left[ \tan^{-1} \left( \frac{2 \times \frac{1}{5}}{1 - (\frac{1}{5})^2} \right) \right] \) (using \( 2\tan^{-1} x = \tan^{-1} \frac{2x}{1-x^2} \))
\( = \tan \left[ \tan^{-1} \left( \frac{2/5}{24/25} \right) \right] = \tan \left[ \tan^{-1} \left( \frac{2}{5} \times \frac{25}{24} \right) \right] = \tan \left[ \tan^{-1} \frac{5}{12} \right] = \frac{5}{12} \)

Question. Write the principal value of \( \cos^{-1} \left( \cos \frac{7\pi}{6} \right) \).
Answer: \( \cos^{-1} (\cos \frac{7\pi}{6}) = \cos^{-1} (\cos (2\pi - \frac{5\pi}{6})) = \cos^{-1} (\cos \frac{5\pi}{6}) = \frac{5\pi}{6} \). (since \( \frac{5\pi}{6} \in [0, \pi] \))

Question. Find the value of \( \sin^{-1} \left( \sin \frac{4\pi}{5} \right) \).
Answer: We are given \( \sin^{-1} (\sin \frac{4\pi}{5}) = \sin^{-1} (\sin (\pi - \frac{\pi}{5})) = \sin^{-1} (\sin \frac{\pi}{5}) = \frac{\pi}{5} \)

Question. Write the principal value of \( \cos^{-1} \left( \frac{1}{2} \right) + 2 \sin^{-1} \left( \frac{1}{2} \right) \). 
Answer: We have, \( \cos^{-1} (\frac{1}{2}) = \cos^{-1} (\cos \frac{\pi}{3}) = \frac{\pi}{3} \). (since \( \frac{\pi}{3} \in [0, \pi] \))
Also, \( \sin^{-1} (\frac{1}{2}) = \sin^{-1} (\sin \frac{\pi}{6}) = \frac{\pi}{6} \). (since \( \frac{\pi}{6} \in [-\frac{\pi}{2}, \frac{\pi}{2}] \))
\( \therefore \cos^{-1} (\frac{1}{2}) + 2 \sin^{-1} (\frac{1}{2}) = \frac{\pi}{3} + 2(\frac{\pi}{6}) = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3} \)

Question. Write the value of \( \cot(\tan^{-1} a + \cot^{-1} a) \). 
Answer: \( \cot(\tan^{-1} a + \cot^{-1} a) = \cot \frac{\pi}{2} = 0 \) [\( \because \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \forall x \in R \)]

Question. Write the value of \( \sin(2 \sin^{-1} \frac{3}{5}) \).
Answer: Let \( \sin^{-1} \frac{3}{5} = y \)
\( \Rightarrow 2 \sin^{-1} \frac{3}{5} = \sin^{-1} \{ 2 \times \frac{3}{5} \sqrt{1 - \frac{9}{25}} \} \) [\( \because 2 \sin^{-1} x = \sin^{-1} (2x\sqrt{1-x^2}) \)]
\( \Rightarrow \sin^{-1} \{ \frac{6}{5} \times \frac{4}{5} \} = \sin^{-1} (\frac{24}{25}) \)
\( \Rightarrow y = \sin^{-1} (\frac{24}{25}) \)
\( \Rightarrow \sin(2 \sin^{-1} \frac{3}{5}) = \frac{24}{25} \)

Question. Write the principal value of \( \tan^{-1} (\tan \frac{7\pi}{6}) \). 
Answer: \( \tan^{-1} (\tan \frac{7\pi}{6}) = \tan^{-1} (\tan (\pi + \frac{\pi}{6})) \)
\( = \tan^{-1} (\tan \frac{\pi}{6}) = \frac{\pi}{6} \) [\( \because \frac{\pi}{6} \in (-\frac{\pi}{2}, \frac{\pi}{2}) \)]

Question. If \( \sin(\sin^{-1} \frac{1}{5} + \cos^{-1} x) = 1 \), then find the value of \( x \). 
Answer: Given \( \sin(\sin^{-1} \frac{1}{5} + \cos^{-1} x) = 1 \)
\( \Rightarrow \sin^{-1} \frac{1}{5} + \cos^{-1} x = \sin^{-1} 1 \)
\( \Rightarrow \sin^{-1} \frac{1}{5} + \cos^{-1} x = \frac{\pi}{2} \)
\( \Rightarrow \sin^{-1} \frac{1}{5} = \frac{\pi}{2} - \cos^{-1} x \)
\( \Rightarrow \sin^{-1} \frac{1}{5} = \sin^{-1} x \)
\( \Rightarrow x = \frac{1}{5} \)

Question. Find the value of \( \sin^{-1}(\cos \frac{43\pi}{5}) \). 
Answer: \( \sin^{-1}(\cos (8\pi + \frac{3\pi}{5})) = \sin^{-1}(\cos \frac{3\pi}{5}) = \sin^{-1}(\sin(\frac{\pi}{2} - \frac{3\pi}{5})) \)
\( = \sin^{-1}(\sin(-\frac{\pi}{10})) = -\frac{\pi}{10} \) [\( \because -\frac{\pi}{10} \in [-\frac{\pi}{2}, \frac{\pi}{2}] \)]

Question. Find the principal value of \( \cos^{-1} [\cos (-680^{\circ})] \). 
Answer: \( \cos^{-1} [\cos (-680^{\circ})] = \cos^{-1} [\cos (680^{\circ})] \) [\( \because \cos (-\theta) = \cos \theta \)]
\( = \cos^{-1} [\cos (720^{\circ} - 40^{\circ})] = \cos^{-1} [\cos (4\pi - 40^{\circ})] = \cos^{-1} (\cos 40^{\circ}) \)
\( = 40^{\circ} \) or \( \frac{2\pi}{9} \) [\( \because 40^{\circ} = \frac{2\pi}{9} \in [0, \pi] \)]

Short Answer Questions-I

Question. Write \( \cot^{-1} \left( \frac{1}{\sqrt{x^2 - 1}} \right), |x| > 1 \) in simplest form. 
Answer: \( \cot^{-1} \left( \frac{1}{\sqrt{x^2 - 1}} \right) \)
Let \( x = \sec \theta \Rightarrow \theta = \sec^{-1} x \)
Now, \( \cot^{-1} \left( \frac{1}{\sqrt{\sec^2 \theta - 1}} \right) = \cot^{-1} \left( \frac{1}{\sqrt{\tan^2 \theta}} \right) \)
\( = \cot^{-1} (\frac{1}{\tan \theta}) = \cot^{-1} (\cot \theta) = \theta = \sec^{-1} x \)

Question. Express \( \tan^{-1} \left( \frac{\cos x}{1 - \sin x} \right), -\frac{3\pi}{2} < x < \frac{\pi}{2} \) in the simplest forms. 
Answer: We have, \( \tan^{-1} \left( \frac{\cos x}{1 - \sin x} \right) \)
\( = \tan^{-1} \left[ \frac{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}{\left( \cos \frac{x}{2} - \sin \frac{x}{2} \right)^2} \right] \)
\( = \tan^{-1} \left[ \frac{\left( \cos \frac{x}{2} + \sin \frac{x}{2} \right) \left( \cos \frac{x}{2} - \sin \frac{x}{2} \right)}{\left( \cos \frac{x}{2} - \sin \frac{x}{2} \right)^2} \right] \)
\( = \tan^{-1} \left[ \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}} \right] \)
\( = \tan^{-1} \left[ \frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}} \right] \) [Dividing \( N^r \) and \( D^r \) by \( \cos \frac{x}{2} \) in the bracket]
\( = \tan^{-1} \left[ \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) \right] \)
\( = \frac{\pi}{4} + \frac{x}{2} \)

Question. If \( 2 \tan^{-1}(\cos \theta) = \tan^{-1}(2 \csc \theta) \), then show that \( \theta = \frac{\pi}{4} \). 
Answer: We have, \( 2 \tan^{-1}(\cos \theta) = \tan^{-1}(2 \csc \theta) \)
\( \Rightarrow \tan^{-1} \left( \frac{2 \cos \theta}{1 - \cos^2 \theta} \right) = \tan^{-1} (2 \csc \theta) \) [\( \because 2 \tan^{-1} x = \tan^{-1} \frac{2x}{1-x^2} \)]
\( \Rightarrow \frac{2 \cos \theta}{\sin^2 \theta} = 2 \csc \theta \)
\( \Rightarrow \cot \theta \cdot 2 \csc \theta = 2 \csc \theta \Rightarrow \cot \theta = 1 \)
\( \Rightarrow \cot \theta = \cot \frac{\pi}{4} \Rightarrow \theta = \frac{\pi}{4} \)

Question. Write the value of \( \tan^{-1} [2 \sin (2 \cos^{-1} \frac{\sqrt{3}}{2})] \). 
Answer: \( \tan^{-1} [2 \sin (2 \cos^{-1} \frac{\sqrt{3}}{2})] = \tan^{-1} [2 \sin (2 \times \frac{\pi}{6})] \) [\( \because \cos^{-1} \frac{\sqrt{3}}{2} = \frac{\pi}{6} \)]
\( = \tan^{-1} [2 \sin (\frac{\pi}{3})] = \tan^{-1} [2 \times \frac{\sqrt{3}}{2}] \)
\( = \tan^{-1} (\sqrt{3}) = \frac{\pi}{3} \)

Question. What is the principal value of \( \cos^{-1} (\cos \frac{2\pi}{3}) + \sin^{-1} (\sin \frac{2\pi}{3}) \)? 
Answer: \( \cos^{-1} (\cos \frac{2\pi}{3}) + \sin^{-1} (\sin \frac{2\pi}{3}) = \frac{2\pi}{3} + \sin^{-1} (\sin (\pi - \frac{\pi}{3})) \)
[\( \because \frac{2\pi}{3} \notin [-\frac{\pi}{2}, \frac{\pi}{2}] \)]
\( = \frac{2\pi}{3} + \sin^{-1} (\sin \frac{\pi}{3}) = \frac{2\pi}{3} + \frac{\pi}{3} \)
\( = \frac{3\pi}{3} = \pi \) [\( \because \sin^{-1}(\sin x) = x \) if \( x \in [-\frac{\pi}{2}, \frac{\pi}{2}] \) and \( \cos^{-1}(\cos x) = x \) if \( x \in [0, \pi] \)]

Question. Find the value of \( 4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{239} \). 
Answer: We have, \( 4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{239} = 2 \cdot 2 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{239} \)
\( = 2 \cdot \tan^{-1} \left[ \frac{2 \times \frac{1}{5}}{1 - (\frac{1}{5})^2} \right] - \tan^{-1} \frac{1}{239} \) [\( \because 2 \tan^{-1} x = \tan^{-1} \frac{2x}{1-x^2} \)]
\( = 2 \cdot \tan^{-1} \left[ \frac{2/5}{24/25} \right] - \tan^{-1} \frac{1}{239} = 2 \cdot \tan^{-1} \left[ \frac{2}{5} \times \frac{25}{24} \right] - \tan^{-1} \frac{1}{239} \)
\( = 2 \tan^{-1} \frac{5}{12} - \tan^{-1} \frac{1}{239} = \tan^{-1} \left[ \frac{2 \times \frac{5}{12}}{1 - (\frac{5}{12})^2} \right] - \tan^{-1} \frac{1}{239} \)
\( = \tan^{-1} \frac{120}{119} - \tan^{-1} \frac{1}{239} = \tan^{-1} \left[ \frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119} \times \frac{1}{239}} \right] \)
\( = \tan^{-1} \left( \frac{28561}{28561} \right) = \tan^{-1} (1) = \frac{\pi}{4} \)