Chapter 2: Inverse Trigonometric Functions - Board Exam Notes

Exam Weightage & Blueprint

Total: ~8 Marks

Inverse Trigonometric Functions is a crucial chapter that connects with Calculus (Differentiation & Integration). Questions are highly formula-based and require memorization of domains and ranges.

Question Type	Marks	Frequency	Focus Topic
MCQ	1	Very High	Principal Values, Domain/Range identification
Short Answer (2M)	2	High	Finding principal values, simplification
Short Answer (3M)	3	Very High	Proving identities, simplifying expressions
Long Answer	4-5	Medium	Complex proofs with multiple identities

⏰ Last 24-Hour Checklist

Domain & Range - Master Table ★★★★★

CRITICAL: This table is the foundation of the chapter. You MUST memorize all domains and ranges perfectly. 90% of mistakes happen here!

Principal Value Branches (The Most Important Table!)

Function	Notation	Domain	Range (Principal Branch)
Arc Sine	$\sin^{-1}x$ or $\arcsin x$	$[-1, 1]$	$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Arc Cosine	$\cos^{-1}x$ or $\arccos x$	$[-1, 1]$	$[0, \pi]$
Arc Tangent	$\tan^{-1}x$ or $\arctan x$	$\mathbb{R}$ (all real numbers)	$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
Arc Cotangent	$\cot^{-1}x$ or $\text{arccot } x$	$\mathbb{R}$ (all real numbers)	$(0, \pi)$
Arc Secant	$\sec^{-1}x$ or $\text{arcsec } x$	$\mathbb{R} - (-1, 1)$ or $\|x\| \geq 1$	$[0, \pi] - \left\{\frac{\pi}{2}\right\}$
Arc Cosecant	$\text{cosec}^{-1}x$ or $\text{arccosec } x$	$\mathbb{R} - (-1, 1)$ or $\|x\| \geq 1$	$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}$

Memory Tricks:
• sin, cosec, tan have $-\frac{\pi}{2}$ in their ranges
• cos, sec, cot have $0$ as starting point in their ranges
• sin, cos, sec, cosec have domain restricted to $[-1, 1]$ or $|x| \geq 1$
• tan, cot accept all real numbers as domain

Common Mistake: Confusing $\sin^{-1}x$ with $\frac{1}{\sin x}$. Remember: $\sin^{-1}x$ is inverse function, NOT reciprocal! The reciprocal is $(\sin x)^{-1} = \text{cosec } x$.

Properties & Identities 🔥🔥🔥

1. Complementary Angle Identities

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

2. Negative Angle Identities

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

Pattern: Odd functions (sin, tan) remain odd when inverted. Even function (cos) behaves differently!

3. Reciprocal Identities

$$\text{cosec}^{-1}x = \sin^{-1}\left(\frac{1}{x}\right), \quad |x| \geq 1$$ $$\sec^{-1}x = \cos^{-1}\left(\frac{1}{x}\right), \quad |x| \geq 1$$ $$\cot^{-1}x = \tan^{-1}\left(\frac{1}{x}\right), \quad x > 0$$

4. Composition Properties

$$\sin(\sin^{-1}x) = x, \quad x \in [-1, 1]$$ $$\sin^{-1}(\sin x) = x, \quad x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$ $$\cos(\cos^{-1}x) = x, \quad x \in [-1, 1]$$ $$\cos^{-1}(\cos x) = x, \quad x \in [0, \pi]$$

ALERT: $\sin^{-1}(\sin x) = x$ ONLY when $x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (principal range). Outside this, you need to adjust!

Addition & Subtraction Formulas ★★★★☆

Tangent Addition Formulas (Most Important!)

$$\tan^{-1}x + \tan^{-1}y = \begin{cases} \tan^{-1}\left(\frac{x + y}{1 - xy}\right), & xy < 1 \\[10pt] \pi + \tan^{-1}\left(\frac{x + y}{1 - xy}\right), & xy > 1, x > 0 \\[10pt] -\pi + \tan^{-1}\left(\frac{x + y}{1 - xy}\right), & xy > 1, x < 0 \end{cases}$$

$$\tan^{-1}x - \tan^{-1}y = \begin{cases} \tan^{-1}\left(\frac{x - y}{1 + xy}\right), & xy > -1 \\[10pt] \pi + \tan^{-1}\left(\frac{x - y}{1 + xy}\right), & xy < -1, x > 0 \\[10pt] -\pi + \tan^{-1}\left(\frac{x - y}{1 + xy}\right), & xy < -1, x < 0 \end{cases}$$

Quick Tip: For most Board exam problems, use the first case where $xy < 1$ or $xy > -1$. Complex cases rarely appear!

Double & Triple Angle Formulas

$$2\sin^{-1}x = \sin^{-1}(2x\sqrt{1-x^2}), \quad x \in \left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]$$ $$2\cos^{-1}x = \cos^{-1}(2x^2 - 1), \quad x \in [0, 1]$$ $$2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right), \quad |x| < 1$$ $$3\sin^{-1}x = \sin^{-1}(3x - 4x^3), \quad x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$$

Solved Examples (Board Marking Scheme)

Q1. Find the principal value of $\sin^{-1}\left(-\frac{1}{2}\right)$. (2 Marks)

Step 1: Setup 0.5 Mark

Let $\sin^{-1}\left(-\frac{1}{2}\right) = y$

Then $\sin y = -\frac{1}{2}$

Step 2: Find the angle 1 Mark

We know that $\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$

The range of principal value branch of $\sin^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Since $-\frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Step 3: Answer 0.5 Mark

Therefore, the principal value is $\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$

Q2. Prove that $3\sin^{-1}x = \sin^{-1}(3x - 4x^3)$ for $x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$. (3 Marks)

Step 1: Substitution 1 Mark

Let $\sin^{-1}x = \theta$, then $x = \sin\theta$

We need to prove: $3\theta = \sin^{-1}(3x - 4x^3)$

Step 2: Use trigonometric identity 1 Mark

We know that $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$

Substituting $x = \sin\theta$:

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

Step 3: Apply inverse 1 Mark

Taking $\sin^{-1}$ on both sides:

$3\theta = \sin^{-1}(3x - 4x^3)$

Since $\theta = \sin^{-1}x$, we get:

$3\sin^{-1}x = \sin^{-1}(3x - 4x^3)$

Hence proved.

Q3. Simplify: $\tan^{-1}\left(\frac{\cos x}{1 + \sin x}\right)$, where $-\frac{\pi}{2} < x < \frac{3\pi}{2}$. (3 Marks)

Step 1: Rewrite using half-angle 1 Mark

We know: $\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2}$ and $1 + \sin x = \left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2$

Alternatively, use: $\cos x = 1 - 2\sin^2\frac{x}{2}$ and $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$

Step 2: Simplify numerator and denominator 1 Mark

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

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

$= \frac{\cos\frac{x}{2} - \sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}}$

Step 3: Divide by $\cos\frac{x}{2}$ 1 Mark

$= \frac{1 - \tan\frac{x}{2}}{1 + \tan\frac{x}{2}} = \frac{\tan\frac{\pi}{4} - \tan\frac{x}{2}}{1 + \tan\frac{\pi}{4}\tan\frac{x}{2}}$

$= \tan\left(\frac{\pi}{4} - \frac{x}{2}\right)$

Therefore: $\tan^{-1}\left(\frac{\cos x}{1 + \sin x}\right) = \frac{\pi}{4} - \frac{x}{2}$

Previous Year Questions (PYQs)

2023 (1 Mark MCQ): The value of $\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)$ is:
(A) $\frac{5\pi}{12}$ (B) $\frac{7\pi}{12}$ (C) $\frac{3\pi}{4}$ (D) $\frac{2\pi}{3}$

Solution: $\tan^{-1}(1) = \frac{\pi}{4}$, $\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$, $\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$
Sum = $\frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6} = \frac{3\pi + 8\pi - 2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}$
Answer: (C)

2022 (2 Marks): Find the value of $\sin^{-1}\left(\sin\frac{3\pi}{5}\right)$.

Solution: Since $\frac{3\pi}{5} \notin \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, we cannot directly write $\sin^{-1}(\sin x) = x$.
We know $\sin\frac{3\pi}{5} = \sin\left(\pi - \frac{3\pi}{5}\right) = \sin\frac{2\pi}{5}$
Since $\frac{2\pi}{5} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, we have:
$\sin^{-1}\left(\sin\frac{3\pi}{5}\right) = \frac{2\pi}{5}$

2020 (3 Marks): Prove that $2\tan^{-1}\frac{1}{3} = \tan^{-1}\frac{3}{4}$.

Solution: Using the formula $2\tan^{-1}x = \tan^{-1}\frac{2x}{1-x^2}$:
LHS = $2\tan^{-1}\frac{1}{3} = \tan^{-1}\frac{2 \cdot \frac{1}{3}}{1 - \left(\frac{1}{3}\right)^2}$
$= \tan^{-1}\frac{\frac{2}{3}}{1 - \frac{1}{9}} = \tan^{-1}\frac{\frac{2}{3}}{\frac{8}{9}} = \tan^{-1}\frac{2 \times 9}{3 \times 8} = \tan^{-1}\frac{3}{4}$ = RHS
Hence proved.

2019 (3 Marks): Prove that $\sin^{-1}\frac{3}{5} + \sin^{-1}\frac{8}{17} = \tan^{-1}\frac{77}{36}$.

Hint: Let $\sin^{-1}\frac{3}{5} = \alpha$ and $\sin^{-1}\frac{8}{17} = \beta$.
Find $\cos\alpha$ and $\cos\beta$ using $\cos^2\theta + \sin^2\theta = 1$.
Calculate $\tan\alpha$ and $\tan\beta$, then use $\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$.

Exam Strategy & Mistake Bank

Common Mistakes 🚨

Mistake 1: Writing $\sin^{-1}(\sin x) = x$ without checking if $x$ is in the principal range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Mistake 2: Confusing domain and range. Remember: Domain is INPUT values, Range is OUTPUT values!

Mistake 3: Using wrong formula for $\tan^{-1}x + \tan^{-1}y$ when $xy > 1$ or $xy < -1$. Check the condition first!

Mistake 4: Forgetting the $\pi$ adjustment in $\cos^{-1}(-x) = \pi - \cos^{-1}x$.

Mistake 5: Writing $\text{cosec}^{-1}x = \sin^{-1}\frac{1}{x}$ without checking $|x| \geq 1$.

Scoring Tips 🏆

Tip 1: Always write the range you're considering. For example: "Since the principal range of $\sin^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$..."

Tip 2: For simplification problems, convert everything to $\tan^{-1}$ form - it's usually easier to work with.

Tip 3: Draw a right triangle for visualization. If $\sin^{-1}x = \theta$, draw a triangle with opposite = $x$ and hypotenuse = $1$.

Tip 4: Memorize standard values: $\sin^{-1}\frac{1}{2} = \frac{\pi}{6}$, $\cos^{-1}\frac{1}{2} = \frac{\pi}{3}$, $\tan^{-1}1 = \frac{\pi}{4}$.

Tip 5: Practice "out of range" questions like $\sin^{-1}(\sin\frac{3\pi}{5})$ - they're common in exams!

Standard Values (Must Memorize!) 📝

Sine Inverse Values

$x$	$\sin^{-1}x$
$-1$	$-\frac{\pi}{2}$
$-\frac{\sqrt{3}}{2}$	$-\frac{\pi}{3}$
$-\frac{1}{\sqrt{2}}$	$-\frac{\pi}{4}$
$-\frac{1}{2}$	$-\frac{\pi}{6}$
$0$	$0$
$\frac{1}{2}$	$\frac{\pi}{6}$
$\frac{1}{\sqrt{2}}$	$\frac{\pi}{4}$
$\frac{\sqrt{3}}{2}$	$\frac{\pi}{3}$
$1$	$\frac{\pi}{2}$

Cosine Inverse Values

$x$	$\cos^{-1}x$
$-1$	$\pi$
$-\frac{\sqrt{3}}{2}$	$\frac{5\pi}{6}$
$-\frac{1}{\sqrt{2}}$	$\frac{3\pi}{4}$
$-\frac{1}{2}$	$\frac{2\pi}{3}$
$0$	$\frac{\pi}{2}$
$\frac{1}{2}$	$\frac{\pi}{3}$
$\frac{1}{\sqrt{2}}$	$\frac{\pi}{4}$
$\frac{\sqrt{3}}{2}$	$\frac{\pi}{6}$
$1$	$0$

Tangent Inverse Values

$x$	$-\sqrt{3}$	$-1$	$-\frac{1}{\sqrt{3}}$	$0$	$\frac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$
$\tan^{-1}x$	$-\frac{\pi}{3}$	$-\frac{\pi}{4}$	$-\frac{\pi}{6}$	$0$	$\frac{\pi}{6}$	$\frac{\pi}{4}$	$\frac{\pi}{3}$

Practice Problems (Self-Assessment)

Level 1: Basic (2 Marks Each)

Q1. Find the principal value of $\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$.

Q2. Find the value of $\tan^{-1}(\sqrt{3}) + \cot^{-1}(\sqrt{3})$.

Q3. Evaluate: $\sin\left(\cos^{-1}\frac{3}{5}\right)$.

Hint: Draw a right triangle or use $\sin^2\theta + \cos^2\theta = 1$.

Level 2: Intermediate (3 Marks Each)

Q4. Prove that $\sin^{-1}\frac{3}{5} + \cos^{-1}\frac{12}{13} = \sin^{-1}\frac{56}{65}$.

Q5. Simplify: $\tan^{-1}\frac{1 - x}{1 + x}$, where $-1 < x < 1$.

Hint: Put $x = \tan\theta$.

Q6. Find the value of $\sin^{-1}(\sin 10)$.

Hint: 10 radians is NOT in the principal range!

Level 3: Advanced (4-5 Marks Each)

Q7. Solve: $\tan^{-1}\frac{x-1}{x-2} + \tan^{-1}\frac{x+1}{x+2} = \frac{\pi}{4}$.

Q8. Prove that $\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{8} = \frac{\pi}{4}$.

Hint: Use addition formula twice, first combine first two terms.

Formula Sheet (Must Remember!) 📋

Principal Value Ranges

1. $\sin^{-1}x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, domain: $[-1, 1]$
2. $\cos^{-1}x \in [0, \pi]$, domain: $[-1, 1]$
3. $\tan^{-1}x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, domain: $\mathbb{R}$
4. $\cot^{-1}x \in (0, \pi)$, domain: $\mathbb{R}$
5. $\sec^{-1}x \in [0, \pi] - \{\frac{\pi}{2}\}$, domain: $|x| \geq 1$
6. $\text{cosec}^{-1}x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}$, domain: $|x| \geq 1$

Complementary Identities

7. $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$
8. $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$
9. $\sec^{-1}x + \text{cosec}^{-1}x = \frac{\pi}{2}$

Negative Angle Identities

10. $\sin^{-1}(-x) = -\sin^{-1}x$
11. $\tan^{-1}(-x) = -\tan^{-1}x$
12. $\cos^{-1}(-x) = \pi - \cos^{-1}x$

Addition Formulas

13. $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ when $xy < 1$
14. $\tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$ when $xy > -1$
15. $2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$ when $|x| < 1$