Exam Weightage & Blueprint
Total: ~4-6 MarksThis 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 (The Two Families)
1. The "Sin Family" (Centered at 0)
Members: $\sin^{-1}, \tan^{-1}, \csc^{-1}$
Range: Roughly $[-\frac{\pi}{2}, \frac{\pi}{2}]$
- $\sin^{-1}$: Closed Interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$
- $\tan^{-1}$: Open Interval $(-\frac{\pi}{2}, \frac{\pi}{2})$
- $\csc^{-1}$: Closed minus 0 ($[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$)
$\sin^{-1}(-x) = -\sin^{-1}x$
2. The "Cos Family" (0 to $\pi$)
Members: $\cos^{-1}, \cot^{-1}, \sec^{-1}$
Range: Roughly $[0, \pi]$
- $\cos^{-1}$: Closed Interval $[0, \pi]$
- $\cot^{-1}$: Open Interval $(0, \pi)$
- $\sec^{-1}$: Closed minus $\pi/2$ ($[0, \pi] - \{\frac{\pi}{2}\}$)
$\cos^{-1}(-x) = \pi - \cos^{-1}x$
| Function | Domain (Input) | Range (Principal Output) |
|---|---|---|
| $\sin^{-1}x$ | $[-1, 1]$ | $[-\frac{\pi}{2}, \frac{\pi}{2}]$ |
| $\cos^{-1}x$ | $[-1, 1]$ | $[0, \pi]$ |
| $\tan^{-1}x$ | $\mathbb{R}$ | $(-\frac{\pi}{2}, \frac{\pi}{2})$ |
| $\cot^{-1}x$ | $\mathbb{R}$ | $(0, \pi)$ |
| $\sec^{-1}x$ | $\mathbb{R} - (-1, 1)$ | $[0, \pi] - \{\frac{\pi}{2}\}$ |
| $\csc^{-1}x$ | $\mathbb{R} - (-1, 1)$ | $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$ |
*Table simplified to most common functions for exams.
Quick Check: Find the principal value of $\sin^{-1}(-\frac{1}{2})$.
Properties & Shortcuts
Group 1: The "Spitters"
Cue: "Just spit the minus out!"
They behave like odd functions.
- $\sin^{-1}(-x) = -\sin^{-1}x$
- $\tan^{-1}(-x) = -\tan^{-1}x$
- $\csc^{-1}(-x) = -\csc^{-1}x$
Group 2: The "$\pi$-Protectors"
Cue: "Protect with $\pi$"
They need $\pi$ to handle the negative.
- $\cos^{-1}(-x) = \pi - \cos^{-1}x$
- $\cot^{-1}(-x) = \pi - \cot^{-1}x$
- $\sec^{-1}(-x) = \pi - \sec^{-1}x$
How to Find Principal Value (Step-by-Step)
- Step 1: check the Sign.
Is it negative? e.g., $\sin^{-1}(-1/2)$.
Use the Spitter or Protector rule to make it positive. - Step 2: Find the Angle.
Ask: "which angle gives this value?"
e.g., for $\sin^{-1}(1/2)$, think $\sin(30^\circ) = 1/2$.
So angle is $30^\circ$ or $\pi/6$. - Step 3: Combine.
Apply the sign from Step 1 to the angle from Step 2.
e.g., $\sin^{-1}(-1/2) = -(\pi/6) = -\pi/6$.
2. Reciprocal Arguments
$\cos^{-1}(\frac{1}{x}) = \sec^{-1}x, \quad x \ge 1 \text{ or } x \le -1$
$\tan^{-1}(\frac{1}{x}) = \cot^{-1}x, \quad x > 0$
3. Complementary Angles
$\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}, \quad x \in \mathbb{R}$
$\sec^{-1}x + \csc^{-1}x = \frac{\pi}{2}, \quad |x| \ge 1$
4. Sum and Difference Formulas
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$
6. Standard Substitutions
- For $\sqrt{a^2 - x^2}$, put $x = a\sin\theta$ or $a\cos\theta$
- For $\sqrt{a^2 + x^2}$, put $x = a\tan\theta$ or $a\cot\theta$
- For $\sqrt{x^2 - a^2}$, put $x = a\sec\theta$ or $a\csc\theta$
- For $\sqrt{\frac{a-x}{a+x}}$ or $\sqrt{\frac{a+x}{a-x}}$, put $x = a\cos 2\theta$
Solved Examples (Board Marking Scheme)
Q1. Find the principal value of $\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2)$. (2 Marks)
$\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$ (since $\tan \frac{\pi}{3} = \sqrt{3}$)
$\sec^{-1}(-2) = \pi - \sec^{-1}(2)$ (using property)
$= \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
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)
Put $x = \tan\theta \Rightarrow \theta = \tan^{-1}x$.
Expression becomes $\tan^{-1}\left(\frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta}\right) = \tan^{-1}\left(\frac{\sec\theta-1}{\tan\theta}\right)$.
$= \tan^{-1}\left(\frac{\frac{1}{\cos\theta}-1}{\frac{\sin\theta}{\cos\theta}}\right) = \tan^{-1}\left(\frac{1-\cos\theta}{\sin\theta}\right)$
Using half-angle formulas: $1-\cos\theta = 2\sin^2\frac{\theta}{2}$ and $\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$.
$= \tan^{-1}\left(\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right) = \tan^{-1}\left(\tan\frac{\theta}{2}\right)$.
$= \frac{\theta}{2} = \frac{1}{2}\tan^{-1}x$.
Previous Year Questions (PYQs)
(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 $\sin^{-1}(\sin\frac{-\pi}{10}) = \frac{-\pi}{10}$.
Ans: Let $\tan^{-1}x = \theta \Rightarrow \tan\theta = \frac{x}{1}$. In right $\Delta$, $P=x, B=1 \Rightarrow H=\sqrt{1+x^2}$.
$\sin\theta = \frac{P}{H} = \frac{x}{\sqrt{1+x^2}}$.
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
Scoring Tips
Practice Problems (Self-Assessment)
Level 1: Basic (1 Mark Each)
Q1. Find the principal value of $\cos^{-1}(-\frac{1}{2})$.
It's negative inside $\cos^{-1}$. Use $\pi - \dots$
$\cos^{-1}(-\frac{1}{2}) = \pi - \cos^{-1}(\frac{1}{2})$
Which angle has $\cos = 1/2$? It is $60^\circ$ or $\frac{\pi}{3}$.
$= \pi - \frac{\pi}{3} = \frac{3\pi - \pi}{3} = \frac{2\pi}{3}$.
Q2. Evaluate $\tan^{-1}(1) + \cos^{-1}(-\frac{1}{2}) + \sin^{-1}(-\frac{1}{2})$.
1. $\tan^{-1}(1) = \frac{\pi}{4}$
2. $\cos^{-1}(-\frac{1}{2}) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ (Protector)
3. $\sin^{-1}(-\frac{1}{2}) = -\frac{\pi}{6}$ (Spitter)
$= \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6}$
LCM of 4, 3, 6 is 12.
$= \frac{3\pi + 8\pi - 2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}$.
Level 2: Intermediate (2-3 Marks Each)
Q3. Prove that $3\sin^{-1}x = \sin^{-1}(3x - 4x^3), x \in [-\frac{1}{2}, \frac{1}{2}]$.
Put $x = \sin\theta \Rightarrow \theta = \sin^{-1}x$.
RHS $= \sin^{-1}(3\sin\theta - 4\sin^3\theta)$
This matches formula $\sin 3\theta$.
$= \sin^{-1}(\sin 3\theta) = 3\theta$
Substitute back $\theta$: $= 3\sin^{-1}x = \text{LHS}$.
Hence Proved.
Q4. Express $\tan^{-1}(\frac{\cos x}{1-\sin x})$ in simplest form.
$\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2}$
$1 - \sin x = (\cos\frac{x}{2} - \sin\frac{x}{2})^2$
Expression $= \tan^{-1}\left( \frac{(\cos\frac{x}{2}-\sin\frac{x}{2})(\cos\frac{x}{2}+\sin\frac{x}{2})}{(\cos\frac{x}{2}-\sin\frac{x}{2})^2} \right)$
$= \tan^{-1}\left( \frac{\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{x}{2}-\sin\frac{x}{2}} \right)$
Divide Nr and Dr by $\cos\frac{x}{2}$:
$= \tan^{-1}\left( \frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}} \right)$
$= \tan^{-1}\left( \tan(\frac{\pi}{4} + \frac{x}{2}) \right)$
$= \frac{\pi}{4} + \frac{x}{2}$.
Level 3: Advanced (4 Marks Each)
Q5. Solve for $x$: $\tan^{-1}(\frac{x-1}{x-2}) + \tan^{-1}(\frac{x+1}{x+2}) = \frac{\pi}{4}$.
$\tan^{-1}A + \tan^{-1}B = \tan^{-1}(\frac{A+B}{1-AB})$
$\frac{x-1}{x-2} + \frac{x+1}{x+2} = \frac{(x-1)(x+2) + (x+1)(x-2)}{(x-2)(x+2)}$
Nr: $(x^2+x-2) + (x^2-x-2) = 2x^2 - 4$
Dr: $(x^2-4) - (x^2-1) = -3$
$\frac{2x^2 - 4}{-3} = \tan(\frac{\pi}{4}) = 1$
$2x^2 - 4 = -3 \Rightarrow 2x^2 = 1$
$x^2 = 1/2 \Rightarrow x = \pm \frac{1}{\sqrt{2}}$.