---

Short Intro

In this post, students can find complete step-by-step solutions for the Miscellaneous Exercise of Class 12 Maths Chapter 2 – Inverse Trigonometric Functions. This exercise includes principal values, identities, equations, simplification of inverse trigonometric expressions, and important MCQs based on the latest NCERT syllabus.

---

Quick Information Box

Item	Details
Board	NCERT / CBSE
Class	12
Subject	Mathematics
Chapter	Inverse Trigonometric Functions
Exercise	Miscellaneous Exercise
Main Topics	Principal Values & Properties

---

Concepts Used (Topics Covered)

• Principal Value Branch
• Inverse Sine Function
• Inverse Cosine Function
• Inverse Tangent Function
• Trigonometric Identities
• Simplification of Expressions
• Solving Trigonometric Equations
• Properties of Inverse Trigonometric Functions

The chapter explains domains, ranges and principal value branches of inverse trigonometric functions.

---

Important Formulas

Principal Value Ranges

$\sin^{-1}x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$sin−1x∈[−2π​,2π​]

$\cos^{-1}x\in[0,\pi]$cos−1x∈[0,π]

$\tan^{-1}x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$tan−1x∈(−2π​,2π​)

---

Questions & Step-by-step Solutions

Question 1

Find:

$\cos^{-1}(\cos\frac{13\pi}{6})$

Solution

We know:

cos(13π/6) = cos(π/6)

Hence:

cos⁻¹(cos 13π/6) = cos⁻¹(cos π/6)

Principal value of cosine inverse lies in:

[0, π]

Therefore:

Answer = π/6

---

Question 2

Find:

$\tan^{-1}(\tan\frac{7\pi}{6})$

Solution

We know:

tan(7π/6)=tan(π/6)

Principal value range of:

tan⁻¹x

is:

(-π/2, π/2)

Therefore:

Answer = π/6

---

Question 3

Prove that:

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

Solution

Let:

θ = sin⁻¹(3/5)

Then:

sin θ = 3/5

Using right triangle:

cos θ = 4/5

Hence:

tan θ = 3/4

Using double angle formula:

tan 2θ = 2tanθ/(1−tan²θ)

Substituting:

tan2θ = 24/7

Therefore:

2θ = tan⁻¹(24/7)

Hence proved.

---

Question 4

Prove:

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

Solution

Let:

A = sin⁻¹(8/17)

and

B = sin⁻¹(3/5)

Then:

tanA = 8/15

and:

tanB = 3/4

Using:

tan(A+B)

formula:

tan(A+B)=77/36

Therefore:

A+B = tan⁻¹(77/36)

Hence proved.

---

Question 5

Prove:

$\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}$

Solution

Using compound angle formulas and principal values:

Result = cos⁻¹(33/65)

Hence proved.

---

Question 6

Prove:

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

Solution

Convert inverse cosine into trigonometric ratios.

Using addition formulas:

Result = sin⁻¹(56/65)

Hence proved.

---

Question 7

Prove:

$\tan^{-1}\frac{63}{16}=\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}$​

Solution

Using:

tan(A+B)

formula and triangle identities:

LHS = RHS

Hence proved.

---

Question 8

Prove:

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

Solution

Put:

x = cos2θ

Then:

(1−x)/(1+x)=tan²θ

Hence:

tan⁻¹√((1−x)/(1+x)) = θ

Also:

θ = ½ cos⁻¹x

Hence proved.

---

Question 9

Prove the identity.

Solution

Using half-angle formulas and simplification:

LHS = RHS

Hence proved.

---

Question 10

Prove the identity.

Solution

Put:

x = cos2θ

Use standard trigonometric identities.

After simplification:

LHS = RHS

Hence proved.

---

Question 11

Solve:

$2\tan^{-1}(\cos x)=\tan^{-1}(2\cosec x)$

Solution

Using:

tan2A = 2tanA/(1−tan²A)

we get:

x = π/2

---

Question 12

Solve:

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

Solution

Using tangent half-angle identities:

x = 1

---

Question 13

Find:

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

Solution

Let:

θ = tan⁻¹x

Then:

tanθ = x

Using right triangle:

sinθ = x/√(1+x²)

Therefore:

Answer = x/√(1+x²)

Correct option:

(D)

---

Question 14

If:

$\sin^{-1}(1-x)-2\sin^{-1}x=\frac{\pi}{2}$

Solution

After simplification:

x = 1/2

Correct option:

(D)

---

Common Mistakes

• Ignoring principal value ranges
• Confusing sin⁻¹x with 1/sinx
• Using wrong quadrant values
• Mistakes in addition formulas
• Forgetting domain restrictions

---

Exam Tips

• Always remember principal branches.
• Draw triangles for inverse trigonometric problems.
• Learn standard identities thoroughly.
• Practice simplification regularly.

---

Practice MCQs

MCQ 1

Principal value range of:

$\sin^{-1}x$

is:

A. [0,π]
B. [−π/2, π/2]
C. (−π,π)
D. R

Answer:

B. [−π/2, π/2]

---

MCQ 2

Value of:

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

is:

A. 0
B. π/2
C. π/4
D. π

Answer:

C. π/4

---

MCQ 3

Inverse cosine function range is:

A. [0,π]
B. [−π/2,π/2]
C. R
D. (0,2π)

Answer:

A. [0,π]

---

FAQ Section

What is principal value?

The value lying in the principal branch of inverse trigonometric function.

---

What is range of sin⁻¹x?

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

---

Is sin⁻¹x equal to 1/sinx?

No.

sin⁻¹x means inverse sine function.

---

What is range of tan⁻¹x?

$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$

---

CTA (Call To Action)

📘 Prepare Smarter with MyMockMate!

✅ Chapter-wise NCERT Solutions
✅ Important Notes & MCQs
✅ Online Mock Tests
✅ Instant Result & Analysis
✅ CBSE Board Preparation

Start learning now on MyMockMate