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Short Intro

In this post, students can find complete step-by-step solutions for Class 12 Maths Chapter 2 Exercise 2.2 – Inverse Trigonometric Functions based on the latest NCERT syllabus. This exercise includes inverse trigonometric identities, simplification of expressions, proving formulas, and solving equations in an easy and exam-oriented format.

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Quick Information Box

Item	Details
Board	NCERT / CBSE
Class	12
Subject	Mathematics
Chapter	Inverse Trigonometric Functions
Exercise	2.2
Main Topics	Identities & Simplifications

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Concepts Used (Topics Covered)

• Inverse Trigonometric Identities
• Triple Angle Formula
• Simplification of Expressions
• Principal Value Branch
• tan⁻¹ Identities
• sin⁻¹ and cos⁻¹ Relations
• Solving Trigonometric Equations
• Principal Values

The exercise focuses on proving identities and simplifying inverse trigonometric expressions.

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Important Formulas

Triple Angle Identity

$\sin3\theta=3\sin\theta-4\sin^3\theta$

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Triple Angle Cosine Formula

$\cos3\theta=4\cos^3\theta-3\cos\theta$

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Tangent Addition Formula

$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$​

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Principal Value Range of tan⁻¹x

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

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Questions & Step-by-step Solutions

Question 1

Prove that:

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

Solution

Let:

x = sinθ

Then:

θ = sin⁻¹x

Using triple angle identity:

$\sin3\theta=3\sin\theta-4\sin^3\theta$

Substituting:

sin3θ = 3x − 4x³

Taking inverse sine:

3θ = sin⁻¹(3x−4x³)

Hence:

3sin⁻¹x = sin⁻¹(3x−4x³)

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Question 2

Prove that:

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

Solution

Let:

x = cosθ

Then:

θ = cos⁻¹x

Using triple angle identity:

$\cos3\theta=4\cos^3\theta-3\cos\theta$

Substituting:

cos3θ = 4x³−3x

Taking inverse cosine:

3θ = cos⁻¹(4x³−3x)

Hence proved.

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Question 3

Simplify:

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

Solution

Multiply numerator and denominator appropriately.

After simplification:

= tan⁻¹[x/(√(1+x²)+1)]

Using tangent half-angle identity:

Answer = ½ tan⁻¹x

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Question 4

Simplify:

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

Solution

Using identity:

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

Therefore:

√[(1−cosx)/(1+cosx)] = tan(x/2)

Hence:

Answer = x/2

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Question 5

Simplify:

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

Solution

Divide numerator and denominator by:

cosx

Then:

= tan⁻¹[(1−tanx)/(1+tanx)]

Using formula:

tan(A−B)

Therefore:

Answer = π/4 − x

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Question 6

Simplify:

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

Solution

Put:

x = a sinθ

Then:

√(a²−x²)=a cosθ

Therefore:

tan⁻¹[tanθ]=θ

Hence:

Answer = sin⁻¹(x/a)

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Question 7

Simplify:

$\tan^{-1}\left(\frac{3a^2x-x^3}{a^3-3ax^2}\right)$

Solution

Put:

x = a tanθ

Using triple angle formula:

tan3θ = (3tanθ−tan³θ)/(1−3tan²θ)

Therefore:

Answer = 3tan⁻¹(x/a)

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Question 8

Find the value of:

$\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac12\right)\right]$

Solution

We know:

sin⁻¹(1/2)=π/6

Therefore:

2(π/6)=π/3

Now:

cos(π/3)=1/2

Hence:

2 × 1/2 = 1

Therefore:

Answer = tan⁻¹(1)=π/4

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Question 9

Find the value of the expression.

Solution

Using standard inverse trigonometric identities and simplification:

Answer = ½ tan⁻¹(x+y)

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Question 10

Find:

$\sin^{-1}(\sin\frac{2\pi}{3})$

Solution

We know:

sin(2π/3)=√3/2

Principal value branch of:

sin⁻¹x

is:

[−π/2, π/2]

Therefore:

Answer = π/3

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Question 11

Find:

$\tan^{-1}(\tan\frac{3\pi}{4})$

Solution

We know:

tan(3π/4)=−1

Hence:

tan⁻¹(−1)=−π/4

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Question 12

Find:

Solution

Using standard values:

Answer = π/2

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Question 13

Evaluate:

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

Solution

Since:

cos(7π/6)=−√3/2

Principal value of cosine inverse lies in:

[0, π]

Hence:

Answer = 5π/6

Correct option:

(B)

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Question 14

Evaluate:

Solution

Since:

sin⁻¹(1/2)=π/6

Therefore:

sin(π/3−π/6)=sin(π/6)

Hence:

Answer = 1/2

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Question 15

Evaluate:

$\tan^{-1}(\sqrt3)-\cot^{-1}(-\sqrt3)$

Solution

We know:

tan⁻¹(√3)=π/3

and:

cot⁻¹(−√3)=5π/6

Therefore:

π/3−5π/6

Hence:

Answer = −π/2

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Common Mistakes

• Ignoring principal value ranges
• Confusing inverse with reciprocal
• Wrong use of triple angle formulas
• Errors in simplification
• Forgetting domain conditions

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Exam Tips

• Memorize standard inverse trigonometric values.
• Practice tangent addition identities.
• Always check principal branch.
• Use substitutions carefully.

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Practice MCQs

MCQ 1

Range of:

$\tan^{-1}x$

is:

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

Answer:

B. (−π/2, π/2)

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MCQ 2

Value of:

$\sin^{-1}\frac12$

is:

A. π/6
B. π/4
C. π/3
D. π/2

Answer:

A. π/6

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MCQ 3

Value of:

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

is:

A. π
B. π/2
C. π/4
D. 0

Answer:

C. π/4

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FAQ Section

What is principal value branch?

The restricted range chosen for inverse trigonometric functions.

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What is the range of tan⁻¹x?

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

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Is sin⁻¹x equal to 1/sinx?

No.

sin⁻¹x means inverse sine function.

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Why are domains restricted?

To make trigonometric functions one-one and invertible.

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