Home / Mathematics / Chapter 2 — Inverse Trigonometric Functions
Class 12 Maths · Chapter 2 · Bihar Board

Inverse Trigonometric Functions

Master the principal value branches with visual graphs, step-by-step solutions with marking scheme, MCQs, Assertion-Reasoning, and previous year questions.

6 Inverse Functions Visual Graphs Step Marking
6Inverse Functions
20MCQs
10Assertion-Reason
8PYQs
Section 01

Basic Concepts

Why do we need inverse trigonometric functions?

2.1 Introduction

We know that the inverse of a function $f$, denoted by $f^{-1}$, exists if $f$ is one-one and onto. But trigonometric functions (like $\sin x$, $\cos x$, etc.) are NOT one-one and onto over their natural domains.

So, to define their inverses, we restrict their domains to specific intervals where they become bijective.

Key Idea

Trigonometric functions are periodic — so they repeat values. By restricting the domain to a specific interval, we make them one-one and onto, allowing inverse functions to exist.

2.2 Principal Value Branch

When we restrict the domain to make the function bijective, the resulting inverse function has a specific range called the principal value branch.

Critical Rule

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

Unless specified, always use the principal value branch!

⚠️ Important Warning

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

$\sin^{-1}$ means "inverse sine", NOT "1 divided by sine".

2.3 Domain and Range Table

This table is the heart of this chapter. Memorize it!

FunctionDomainRange (Principal Value)
$y = \sin^{-1}x$$[-1, 1]$$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$y = \cos^{-1}x$$[-1, 1]$$[0, \pi]$
$y = \tan^{-1}x$$\mathbb{R}$$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
$y = \cot^{-1}x$$\mathbb{R}$$(0, \pi)$
$y = \sec^{-1}x$$\mathbb{R} - (-1, 1)$$[0, \pi] - \left\{\frac{\pi}{2}\right\}$
$y = \csc^{-1}x$$\mathbb{R} - (-1, 1)$$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}$
Section 02

🎨 Visual Graphs

Understand inverse functions through their graphs

V1 $y = \sin x$ and $y = \sin^{-1}x$

Mirror Image Across y = x X Y y = x y = sin x y = sin⁻¹x O The two curves are mirror images of each other across the line y = x
To get the graph of $\sin^{-1}x$, reflect $\sin x$ over the line $y = x$. We only keep the branch where $x \in [-1,1]$ and $y \in [-\pi/2, \pi/2]$.

V2 All Principal Value Branches

Ranges at a Glance π/2 0 π -π/2 sin⁻¹, tan⁻¹, csc⁻¹ cos⁻¹, cot⁻¹, sec⁻¹ sin⁻¹, tan⁻¹, csc⁻¹ use angles near 0 cos⁻¹, cot⁻¹, sec⁻¹ use angles near π/2
Left-half functions ($\sin^{-1}, \tan^{-1}, \csc^{-1}$) use negative-to-positive range. Right-half functions ($\cos^{-1}, \cot^{-1}, \sec^{-1}$) use positive range near 0 and π.

V3 $\sin^{-1}$ vs $\cos^{-1}$ — Ranges Compared

Same domain [-1,1], Different Ranges Y X y = sin⁻¹x -π/2 π/2 (-1, -π/2) (1, π/2) Y X y = cos⁻¹x π 0 (-1, π) (1, 0) sin⁻¹ range: [-π/2, π/2] | cos⁻¹ range: [0, π]
Both functions have the same domain $[-1, 1]$, but their principal value ranges are different.

V4 $y = \tan^{-1}x$ — Domain $\mathbb{R}$, Range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Asymptotic Behavior X Y y = π/2 y = -π/2 O x → ∞ x → -∞ y = tan⁻¹x
The graph never touches the horizontal dashed lines $y = \pm\pi/2$. These are horizontal asymptotes — the curve approaches them but never reaches.

V5 Complementary Relation Visualized

$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ sin⁻¹x θ cos⁻¹x π/2 - θ Total π/2 (always constant!) sin⁻¹x + cos⁻¹x = π/2 for all x ∈ [-1, 1]
No matter what $x$ is, $\sin^{-1}x + \cos^{-1}x$ always equals $\frac{\pi}{2}$. This is a complementary relation.
Quick Reference

📐 All Important Formulas

Memorize these for board exams

F1 Complementary Pairs

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

F2 Negative Angle Identities

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

F3 Reciprocal Relations

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

F4 Sum and Difference Formulas

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

If $xy > 1$ and $x, y > 0$, then add $\pi$. If $xy > 1$ and $x, y < 0$, then subtract $\pi$.

F5 Double Angle Formulas

$$2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right), \ |x| < 1$$
$$2\sin^{-1}x = \sin^{-1}\left(2x\sqrt{1-x^2}\right), \ |x| \leq \frac{1}{\sqrt{2}}$$
$$2\cos^{-1}x = \cos^{-1}(2x^2 - 1), \ 0 \leq x \leq 1$$

F6 Triple Angle Formulas

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

F7 Composition Rules

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

⚠️ Important: $\sin^{-1}(\sin x) = x$ ONLY when $x$ is in the principal range. Otherwise, first bring $x$ to the principal range using $\sin(\pi - x) = \sin x$ etc.

Shortcuts

💡 Tricks & Shortcuts

Speed techniques for board exams

Trick 1 — Domain-Range Memory Trick

"Co-functions use positive angle"

  • Co- functions: cos⁻¹, cot⁻¹, sec⁻¹ → Range starts from 0
  • Others: sin⁻¹, tan⁻¹, csc⁻¹ → Range centered around 0
Trick 2 — Negative Input Rule

Only sin⁻¹, tan⁻¹, csc⁻¹ give negative outputs for negative inputs. All others ($\cos^{-1}, \cot^{-1}, \sec^{-1}$) give positive outputs by shifting by π.

Trick 3 — $\sin^{-1}(\sin x)$ Value Check

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

  • If $x \in [-\pi/2, \pi/2]$ → answer is $x$
  • If $x = \pi - a$ where $a \in [-\pi/2, \pi/2]$ → answer is $a$
  • If $x = \pi + a$ → answer is $-a$

Similarly for cos⁻¹, tan⁻¹ etc. with their own ranges.

Trick 4 — Common Values Table

Memorize these standard values:

  • $\sin^{-1}(1/2) = \pi/6$, $\sin^{-1}(1/\sqrt{2}) = \pi/4$, $\sin^{-1}(\sqrt{3}/2) = \pi/3$
  • $\cos^{-1}(1/2) = \pi/3$, $\cos^{-1}(1/\sqrt{2}) = \pi/4$, $\cos^{-1}(\sqrt{3}/2) = \pi/6$
  • $\tan^{-1}(1) = \pi/4$, $\tan^{-1}(\sqrt{3}) = \pi/3$, $\tan^{-1}(1/\sqrt{3}) = \pi/6$
Trick 5 — tan⁻¹ Sum Fast Check

For $\tan^{-1}x + \tan^{-1}y$:

  • If $xy < 1$ → use $\tan^{-1}\left(\frac{x+y}{1-xy}\right)$ directly
  • If $xy > 1$ and $x,y > 0$ → add π
  • If $xy > 1$ and $x,y < 0$ → subtract π
Trick 6 — Substitution Trick

When you see expressions like $\sqrt{1-x^2}$, $\frac{1}{\sqrt{1-x^2}}$, $\frac{x}{\sqrt{1-x^2}}$:

  • $x = \sin\theta$ → $\sqrt{1-x^2} = \cos\theta$
  • $x = \cos\theta$ → $\sqrt{1-x^2} = \sin\theta$
  • $x = \tan\theta$ → $1+x^2 = \sec^2\theta$
Trick 7 — Range Verification

Whenever you get an answer for $\sin^{-1}$, $\cos^{-1}$, etc., always check:

  • Is your answer inside the principal value range?
  • If not, adjust by adding/subtracting π
  • For $\sin^{-1}$, cos⁻¹, the answer MUST be in $[-\pi/2, \pi/2]$ or $[0, \pi]$
Trick 8 — π + Angle Reduction

Common reductions:

  • $\sin^{-1}(\sin(\pi - \theta)) = \theta$ (if $\theta \in [-\pi/2, \pi/2]$)
  • $\cos^{-1}(\cos(\pi + \theta)) = \pi - \theta$ (adjust range)
  • $\tan^{-1}(\tan(\pi - \theta)) = -\theta$
Trick 9 — Reciprocal Conversion

Fast reciprocal:

  • $\csc^{-1}(2) = \sin^{-1}(1/2) = \pi/6$
  • $\sec^{-1}(2) = \cos^{-1}(1/2) = \pi/3$
  • $\cot^{-1}(\sqrt{3}) = \tan^{-1}(1/\sqrt{3}) = \pi/6$
Trick 10 — "π minus" trick for negatives

If input is negative and function is cos⁻¹, cot⁻¹, sec⁻¹:

  • Compute the positive version first
  • Then use: $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$
  • Example: $\cos^{-1}(-1/2) = \pi - \cos^{-1}(1/2) = \pi - \pi/3 = 2\pi/3$
Practice

📝 Solved Examples with Step Marking

Click "Show Answer" to reveal step-by-step solutions with marks

📋 Total: 2 Marks · Board Pattern

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

Step 1 — Setup 0.5 Mark

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

Step 2 — Apply Principal Range 1 Mark

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

We know $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and $\frac{\pi}{4} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Step 3 — Final Answer 0.5 Mark

Therefore, $y = \frac{\pi}{4}$.

Principal value of $\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$
📋 Total: 2 Marks

Q2. Find the principal value of $\cot^{-1}\left(\frac{-1}{\sqrt{3}}\right)$.

Step 1 — Setup 0.5 Mark

Let $\cot^{-1}\left(\frac{-1}{\sqrt{3}}\right) = y$. Then $\cot y = \frac{-1}{\sqrt{3}}$.

Step 2 — Use negative angle identity 0.75 Mark

$\cot y = -\cot\left(\frac{\pi}{3}\right) = \cot\left(\pi - \frac{\pi}{3}\right) = \cot\left(\frac{2\pi}{3}\right)$

Step 3 — Check principal range 0.75 Mark

Principal value branch of $\cot^{-1}$ is $(0, \pi)$.

$\frac{2\pi}{3} \in (0, \pi)$ ✓

Principal value of $\cot^{-1}\left(\frac{-1}{\sqrt{3}}\right) = \frac{2\pi}{3}$
📋 Total: 3 Marks

Q3. Show that $\sin^{-1}\left(2x\sqrt{1-x^2}\right) = 2\sin^{-1}x$ for $-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$.

Step 1 — Substitution 1 Mark

Let $x = \sin\theta$. Then $\sin^{-1}x = \theta$, where $\theta \in \left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$.

Step 2 — Simplify LHS 1 Mark

$\sin^{-1}\left(2x\sqrt{1-x^2}\right) = \sin^{-1}\left(2\sin\theta\sqrt{1-\sin^2\theta}\right)$

$= \sin^{-1}(2\sin\theta \cos\theta) = \sin^{-1}(\sin 2\theta) = 2\theta$

Step 3 — Substitute back 1 Mark

Since $2\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, $\sin^{-1}(\sin 2\theta) = 2\theta$.

$= 2\sin^{-1}x$.

$\sin^{-1}\left(2x\sqrt{1-x^2}\right) = 2\sin^{-1}x$ ✓
Test Yourself

✍️ MCQ Practice — 20 Questions

Select your answer and click "Check Answers"

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

Correct: B — $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$, and $\pi/6 \in [-\pi/2, \pi/2]$.

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

Correct: C — The principal value branch of $\cos^{-1}$ is $[0, \pi]$.

3. $\sin^{-1}x + \cos^{-1}x$ equals:

Correct: A — Fundamental complementary relation, valid for all $x \in [-1,1]$.

4. $\tan^{-1}(-1)$ equals:

Correct: D — $\tan^{-1}(-x) = -\tan^{-1}(x)$, so $\tan^{-1}(-1) = -\pi/4$.

5. The domain of $\tan^{-1}x$ is:

Correct: B — Domain of $\tan^{-1}$ is the entire set of real numbers $\mathbb{R}$.

6. $\cos^{-1}\left(-\frac{1}{2}\right)$ equals:

Correct: C — $\cos^{-1}(-x) = \pi - \cos^{-1}x = \pi - \pi/3 = 2\pi/3$.

7. $\tan^{-1}(\sqrt{3})$ equals:

Correct: A — $\tan(\pi/3) = \sqrt{3}$.

8. $\sin^{-1}(\sin(2\pi/3))$ equals:

Correct: C — $\sin(2\pi/3) = \sin(\pi - \pi/3) = \sin(\pi/3)$. Since $\pi/3 \in [-\pi/2, \pi/2]$, $\sin^{-1}(\sin(2\pi/3)) = \pi/3$.

9. $\sec^{-1}(2)$ equals:

Correct: B — $\sec^{-1}(2) = \cos^{-1}(1/2) = \pi/3$.

10. The principal value of $\csc^{-1}(2)$ is:

Correct: A — $\csc^{-1}(2) = \sin^{-1}(1/2) = \pi/6$.

11. The value of $\tan^{-1}(1) + \cos^{-1}(-1/2) + \sin^{-1}(-1/2)$ is:

Correct: C — $\tan^{-1}(1) = \pi/4$, $\cos^{-1}(-1/2) = 2\pi/3$, $\sin^{-1}(-1/2) = -\pi/6$. Sum $= \pi/4 + 2\pi/3 - \pi/6 = 3\pi/12 + 8\pi/12 - 2\pi/12 = 9\pi/12 = 3\pi/4$.

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

Correct: B — Range of $\sin^{-1}x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

13. $\tan^{-1}\sqrt{3} - \sec^{-1}(-2)$ equals:

Correct: D — $\tan^{-1}\sqrt{3} = \pi/3$. $\sec^{-1}(-2) = \pi - \sec^{-1}(2) = \pi - \pi/3 = 2\pi/3$. Difference $= \pi/3 - 2\pi/3 = -\pi/3$.

14. $\cos^{-1}\left(\cos\frac{7\pi}{6}\right)$ equals:

Correct: A — $\cos(7\pi/6) = \cos(\pi + \pi/6) = -\cos(\pi/6) = -\sqrt{3}/2$. $\cos^{-1}(-\sqrt{3}/2) = \pi - \pi/6 = 5\pi/6$.

15. The value of $\sin\left(\frac{\pi}{3} - \sin^{-1}(-1/2)\right)$ is:

Correct: C — $\sin^{-1}(-1/2) = -\pi/6$. So $\pi/3 - (-\pi/6) = \pi/3 + \pi/6 = \pi/2$. $\sin(\pi/2) = 1$.

16. $\tan^{-1}\sqrt{3} - \cot^{-1}(-\sqrt{3})$ equals:

Correct: B — $\tan^{-1}\sqrt{3} = \pi/3$. $\cot^{-1}(-\sqrt{3}) = \pi - \pi/6 = 5\pi/6$. Difference = $\pi/3 - 5\pi/6 = -\pi/2$.

17. $\sin(\tan^{-1}x)$ for $|x| < 1$ equals:

Correct: A — Let $\tan^{-1}x = \theta$, so $\tan\theta = x$. Then $\sin\theta = \frac{x}{\sqrt{1+x^2}}$.

18. If $\sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2}$, then $x$ equals:

Correct: C — Try $x = 0$: $\sin^{-1}(1) - 0 = \pi/2$ ✓. Try $x = 1/2$: $\sin^{-1}(1/2) - 2\sin^{-1}(1/2) = \pi/6 - 2\pi/6 = -\pi/6 \neq \pi/2$. So only $x = 0$.

19. The value of $\sin^{-1}(\sin(3\pi/5))$ is:

Correct: D — $\sin(3\pi/5) = \sin(\pi - 3\pi/5) = \sin(2\pi/5)$. And $2\pi/5 \in [-\pi/2, \pi/2]$. So answer is $2\pi/5$.

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

Correct: B — $\cos(13\pi/6) = \cos(2\pi + \pi/6) = \cos(\pi/6)$. So $\cos^{-1}(\cos(13\pi/6)) = \pi/6$.
Score: 0 / 20
New Pattern

🎯 Assertion & Reasoning — 10 Questions

Each question has an Assertion (A) and a Reason (R). Choose the correct option.

📖 Standard Options

(A) Both A and R are true, and R is the correct explanation of A.

(B) Both A and R are true, but R is NOT the correct explanation of A.

(C) A is true, but R is false.

(D) A is false, but R is true.

1
Assertion (A) The value of $\sin^{-1}\left(\frac{1}{2}\right)$ is $\frac{\pi}{6}$.
Reason (R) The principal value branch of $\sin^{-1}$ has range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $\sin(\pi/6) = 1/2$.
Correct: A — Both A and R are true. R correctly explains why A is true.
2
Assertion (A) $\cos^{-1}(-x) = -\cos^{-1}(x)$ for all $x \in [-1, 1]$.
Reason (R) $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$.
Correct: D — A is false (the correct identity is $\cos^{-1}(-x) = \pi - \cos^{-1}x$), but R is true (this is the actual identity).
3
Assertion (A) $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for all $x \in [-1, 1]$.
Reason (R) $\sin^{-1}x$ and $\cos^{-1}x$ are complementary inverse trigonometric functions.
Correct: A — Both true. R correctly explains why the sum is π/2.
4
Assertion (A) $\tan^{-1}(-1) = \frac{\pi}{4}$.
Reason (R) $\tan^{-1}(-x) = -\tan^{-1}(x)$ for all $x \in \mathbb{R}$.
Correct: D — A is false ($\tan^{-1}(-1) = -\pi/4$, not $\pi/4$). R is true (this is the correct identity).
5
Assertion (A) The domain of $\sec^{-1}x$ is $\mathbb{R} - (-1, 1)$.
Reason (R) $\sec x$ has range $\mathbb{R} - (-1, 1)$.
Correct: A — Both true. The domain of an inverse function equals the range of the original function.
6
Assertion (A) $\sin^{-1}x$ is defined for all real values of $x$.
Reason (R) The domain of $\sin^{-1}x$ is $[-1, 1]$.
Correct: D — A is false: $\sin^{-1}x$ is defined only for $x \in [-1, 1]$, not for all real values. R is true (the domain is indeed $[-1, 1]$).
7
Assertion (A) $\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$.
Reason (R) The principal value branch of $\cos^{-1}$ is $[0, \pi]$ and $\cos(\pi/3) = 1/2$.
Correct: A — Both true. R correctly explains A.
8
Assertion (A) $\sin^{-1}(\sin x) = x$ for all $x \in \mathbb{R}$.
Reason (R) $\sin(\sin^{-1}x) = x$ for all $x \in [-1, 1]$.
Correct: D — A is false ($\sin^{-1}(\sin x) = x$ only when $x \in [-\pi/2, \pi/2]$). R is true.
9
Assertion (A) $\csc^{-1}(2) = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$.
Reason (R) $\csc^{-1}x = \sin^{-1}\left(\frac{1}{x}\right)$ for $|x| \geq 1$.
Correct: A — Both true, and R directly explains A.
10
Assertion (A) $\cot^{-1}(\sqrt{3}) = \frac{\pi}{6}$.
Reason (R) $\cot^{-1}x = \tan^{-1}(1/x)$ for $x > 0$.
Correct: A — Both true. $\cot^{-1}(\sqrt{3}) = \tan^{-1}(1/\sqrt{3}) = \pi/6$, and R is the identity that justifies this.
Score: 0 / 10
Exam Focus

📖 Previous Year Questions

Frequently asked questions in board exams

2024

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

Step 10.5 Mark

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

Step 21 Mark

$\sin y = -\frac{1}{2} = -\sin\left(\frac{\pi}{6}\right) = \sin\left(-\frac{\pi}{6}\right)$.

Step 30.5 Mark

Since $-\pi/6 \in [-\pi/2, \pi/2]$, principal value = $-\pi/6$.

Principal value = $-\frac{\pi}{6}$
2024

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

$\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$ and $\pi/6 \in [0, \pi]$.

$\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$
2023

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

Step 1 — Substitution1 Mark

Let $x = \sin\theta$. Then $\sin^{-1}x = \theta$, where $\theta \in [-\pi/6, \pi/6]$.

Step 2 — RHS simplification1.5 Marks

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

Step 3 — Back substitution0.5 Mark

$= 3\sin^{-1}x$. Hence proved.

$3\sin^{-1}x = \sin^{-1}(3x - 4x^3)$ ✓
2023

Q. Write $\tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right)$ in simplest form, $-\pi/4 < x < 3\pi/4$.

Step 1 — Divide by cos x1 Mark

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

Step 2 — Use tan(A-B)1 Mark

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

Step 3 — Check range0.5 Mark

$\tan^{-1}\left[\tan\left(\frac{\pi}{4} - x\right)\right] = \frac{\pi}{4} - x$ (checking range)

$\frac{\pi}{4} - x$
2022

Q. Find the value of $\cos^{-1}\left(\cos\frac{13\pi}{6}\right)$.

$\cos(13\pi/6) = \cos(2\pi + \pi/6) = \cos(\pi/6)$

Since $\pi/6 \in [0, \pi]$, $\cos^{-1}(\cos(13\pi/6)) = \pi/6$.

$\frac{\pi}{6}$
2022

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

Step 1 — Substitute1 Mark

Let $\sin^{-1}(3/5) = \theta$. Then $\sin\theta = 3/5$ and $\cos\theta = 4/5$.

Step 2 — Compute sin 2θ1 Mark

$\sin 2\theta = 2\sin\theta\cos\theta = 2(3/5)(4/5) = 24/25$

$\cos 2\theta = 1 - 2\sin^2\theta = 1 - 18/25 = 7/25$

Step 3 — tan 2θ1 Mark

$\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{24/25}{7/25} = 24/7$

So $2\theta = \tan^{-1}(24/7)$, i.e., $2\sin^{-1}(3/5) = \tan^{-1}(24/7)$. ✓

Proved ✓
2021

Q. $\sin^{-1}(1-x) - 2\sin^{-1}x = \pi/2$, then $x$ equals:

Try $x = 0$: LHS $= \sin^{-1}(1) - 2\sin^{-1}(0) = \pi/2 - 0 = \pi/2$. ✓

Try $x = 1/2$: LHS $= \sin^{-1}(1/2) - 2\sin^{-1}(1/2) = -\pi/6 \neq \pi/2$. ✗

$x = 0$
2020

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

Step 1 — Setup1 Mark

Let $\sin^{-1}(8/17) = A$ and $\sin^{-1}(3/5) = B$.

So $\sin A = 8/17$, $\cos A = 15/17$; $\sin B = 3/5$, $\cos B = 4/5$.

Step 2 — Find sin(A+B)1 Mark

$\sin(A+B) = \sin A\cos B + \cos A\sin B = \frac{8}{17}\cdot\frac{4}{5} + \frac{15}{17}\cdot\frac{3}{5} = \frac{32}{85} + \frac{45}{85} = \frac{77}{85}$

$\cos(A+B) = \cos A\cos B - \sin A\sin B = \frac{15}{17}\cdot\frac{4}{5} - \frac{8}{17}\cdot\frac{3}{5} = \frac{60}{85} - \frac{24}{85} = \frac{36}{85}$

Step 3 — Find tan(A+B)1 Mark

$\tan(A+B) = \frac{77/85}{36/85} = \frac{77}{36}$

So $A + B = \tan^{-1}(77/36)$, i.e., $\sin^{-1}(8/17) + \sin^{-1}(3/5) = \tan^{-1}(77/36)$. ✓

Proved ✓