Home Class 12 Inverse Trigonometric Functions

Chapter 2: Inverse Trigonometric Functions

Complete Board Exam Focused Notes with Principal Values, Properties & PYQs

Exam Weightage & Blueprint

Total: ~4-6 Marks

This chapter is relatively small but highly scoring. It forms the basis for Differentiation and Integration of inverse trigonometric functions.

Question Type Marks Frequency Focus Topic
MCQ 1 Very High Principal Value Branch, Domain
Short Answer (2M) 2 High Simplification, Properties of ITF
Short Answer (3M) 3 Medium Solving Equations, Proving Identities

Last 24-Hour Checklist

Domain and Range (Principal Value Branch)

Concept: Inverse trigonometric functions are defined only when we restrict the domain of trigonometric functions to make them bijective (one-one and onto). The restricted range is called the Principal Value Branch.
Function Domain Range (Principal Branch)
$y = \sin^{-1}x$ $[-1, 1]$ $[-\frac{\pi}{2}, \frac{\pi}{2}]$
$y = \cos^{-1}x$ $[-1, 1]$ $[0, \pi]$
$y = \tan^{-1}x$ $\mathbb{R}$ $(-\frac{\pi}{2}, \frac{\pi}{2})$
$y = \cot^{-1}x$ $\mathbb{R}$ $(0, \pi)$
$y = \sec^{-1}x$ $\mathbb{R} - (-1, 1)$ $[0, \pi] - \{\frac{\pi}{2}\}$
$y = \csc^{-1}x$ $\mathbb{R} - (-1, 1)$ $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$
Memory Aid:
• Group 1 ($\sin^{-1}, \tan^{-1}, \csc^{-1}$): Range is centered around $0$ (from $-\pi/2$ to $\pi/2$).
• Group 2 ($\cos^{-1}, \cot^{-1}, \sec^{-1}$): Range is from $0$ to $\pi$.

Quick Check: Find the principal value of $\sin^{-1}(-\frac{1}{2})$.

Answer: Since $\sin(\pi/6) = 1/2$, and range is $[-\pi/2, \pi/2]$, $\sin^{-1}(-1/2) = -\pi/6$.

Properties of Inverse Trigonometric Functions

1. Negative Arguments

Group 1 (Odd-like)

$\sin^{-1}(-x) = -\sin^{-1}x$

$\tan^{-1}(-x) = -\tan^{-1}x$

$\csc^{-1}(-x) = -\csc^{-1}x$

Group 2 ($\pi$ minus)

$\cos^{-1}(-x) = \pi - \cos^{-1}x$

$\cot^{-1}(-x) = \pi - \cot^{-1}x$

$\sec^{-1}(-x) = \pi - \sec^{-1}x$

2. Reciprocal Arguments

$\sin^{-1}(\frac{1}{x}) = \csc^{-1}x, \quad x \ge 1 \text{ or } x \le -1$
$\cos^{-1}(\frac{1}{x}) = \sec^{-1}x, \quad x \ge 1 \text{ or } x \le -1$
$\tan^{-1}(\frac{1}{x}) = \cot^{-1}x, \quad x > 0$

3. Complementary Angles

$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}, \quad x \in [-1, 1]$
$\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}, \quad x \in \mathbb{R}$
$\sec^{-1}x + \csc^{-1}x = \frac{\pi}{2}, \quad |x| \ge 1$

4. Sum and Difference Formulas

Sum: $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right), \quad xy < 1$
Difference: $\tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right), \quad xy > -1$

5. Formulas for $2\tan^{-1}x$

In terms of $\sin^{-1}$:

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

$|x| \le 1$

In terms of $\cos^{-1}$:

$\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

$x \ge 0$

In terms of $\tan^{-1}$:

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

$-1 < x < 1$

Solved Examples (Board Marking Scheme)

Q1. Find the principal value of $\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2)$. (2 Marks)

Step 1: Evaluate terms separately 1 Mark

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

$\sec^{-1}(-2) = \pi - \sec^{-1}(2)$ (using property)

$= \pi - \frac{\pi}{3} = \frac{2\pi}{3}$

Step 2: Final Calculation 1 Mark

Expression $= \frac{\pi}{3} - \frac{2\pi}{3}$

$= -\frac{\pi}{3}$

Q2. Write in simplest form: $\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right), x \neq 0$. (3 Marks)

Step 1: Substitution 1 Mark

Put $x = \tan\theta \Rightarrow \theta = \tan^{-1}x$.

Expression becomes $\tan^{-1}\left(\frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta}\right) = \tan^{-1}\left(\frac{\sec\theta-1}{\tan\theta}\right)$.

Step 2: Simplify Trigonometry 1 Mark

$= \tan^{-1}\left(\frac{\frac{1}{\cos\theta}-1}{\frac{\sin\theta}{\cos\theta}}\right) = \tan^{-1}\left(\frac{1-\cos\theta}{\sin\theta}\right)$

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

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

Step 3: Final Answer 1 Mark

$= \frac{\theta}{2} = \frac{1}{2}\tan^{-1}x$.

Previous Year Questions (PYQs)

2023 (1 Mark MCQ): The value of $\sin^{-1}(\cos(\frac{33\pi}{5}))$ is:
(A) $\frac{3\pi}{5}$   (B) $\frac{-7\pi}{5}$   (C) $\frac{\pi}{10}$   (D) $\frac{-\pi}{10}$
Ans: (D). $\cos\frac{33\pi}{5} = \cos(6\pi + \frac{3\pi}{5}) = \cos\frac{3\pi}{5} = \sin(\frac{\pi}{2} - \frac{3\pi}{5}) = \sin(\frac{-\pi}{10})$. So $\sin^{-1}(\sin\frac{-\pi}{10}) = \frac{-\pi}{10}$.
2022 (2 Marks): Find the value of $\sin(\tan^{-1}x)$, where $|x| < 1$.
Ans: Let $\tan^{-1}x = \theta \Rightarrow \tan\theta = \frac{x}{1}$. In right $\Delta$, $P=x, B=1 \Rightarrow H=\sqrt{1+x^2}$.
$\sin\theta = \frac{P}{H} = \frac{x}{\sqrt{1+x^2}}$.
2020 (4 Marks): Solve for $x$: $\tan^{-1}2x + \tan^{-1}3x = \frac{\pi}{4}$.
Hint: Use $\tan^{-1}(\frac{2x+3x}{1-6x^2}) = \frac{\pi}{4} \Rightarrow \frac{5x}{1-6x^2} = 1$. Solve quadratic $6x^2+5x-1=0$. Check $x$ satisfies $xy < 1$.

Exam Strategy & Mistake Bank

Common Mistakes

Mistake 1: Writing $\sin^{-1}x = (\sin x)^{-1} = \frac{1}{\sin x}$. This is WRONG. $\sin^{-1}x$ is the inverse function, not reciprocal.
Mistake 2: Ignoring the domain for formulas. E.g., applying $\tan^{-1}x + \tan^{-1}y$ when $xy > 1$.
Mistake 3: Canceling $\sin^{-1}(\sin \theta) = \theta$ when $\theta$ is outside $[-\pi/2, \pi/2]$. You must bring it into range first.

Scoring Tips

Tip 1: Always mention the Principal Value Branch when solving for principal values.
Tip 2: For simplification problems involving $\sqrt{a^2-x^2}$, put $x = a\sin\theta$. For $\sqrt{a^2+x^2}$, put $x = a\tan\theta$.
Tip 3: When solving equations, always verify the answer by putting it back in the original equation (to check for extraneous roots).

Practice Problems (Self-Assessment)

Level 1: Basic (1 Mark Each)

Q1. Find the principal value of $\cos^{-1}(-\frac{1}{2})$.

Answer: $\pi - \cos^{-1}(1/2) = \pi - \pi/3 = 2\pi/3$.

Q2. Evaluate $\tan^{-1}(1) + \cos^{-1}(-\frac{1}{2}) + \sin^{-1}(-\frac{1}{2})$.

Answer: $\frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6} = \frac{3\pi + 8\pi - 2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}$.

Level 2: Intermediate (2-3 Marks Each)

Q3. Prove that $3\sin^{-1}x = \sin^{-1}(3x - 4x^3), x \in [-\frac{1}{2}, \frac{1}{2}]$.

Hint: Put $x = \sin\theta$. RHS becomes $\sin^{-1}(\sin 3\theta) = 3\theta = 3\sin^{-1}x$.

Q4. Express $\tan^{-1}(\frac{\cos x}{1-\sin x})$ in simplest form.

Answer: $\tan^{-1}(\tan(\frac{\pi}{4} + \frac{x}{2})) = \frac{\pi}{4} + \frac{x}{2}$.

Level 3: Advanced (4 Marks Each)

Q5. Solve for $x$: $\tan^{-1}(\frac{x-1}{x-2}) + \tan^{-1}(\frac{x+1}{x+2}) = \frac{\pi}{4}$.

Answer: Apply $\tan^{-1}A + \tan^{-1}B$ formula. You will get $2x^2 - 4 = -3 \Rightarrow 2x^2 = 1 \Rightarrow x = \pm \frac{1}{\sqrt{2}}$.
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