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Short Intro

In this post, students can find complete step-by-step solutions for Class 12 Maths Chapter 2 Exercise 2.1 – Inverse Trigonometric Functions based on the latest NCERT syllabus. This exercise covers principal values, domains and ranges of inverse trigonometric functions, and evaluation of inverse trigonometric expressions in a simple 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	2.1
Main Topics	Principal Values

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Concepts Used (Topics Covered)

• Principal Value Branch
• Inverse Sine Function
• Inverse Cosine Function
• Inverse Tangent Function
• Inverse Cotangent Function
• Inverse Secant Function
• Inverse Cosecant Function
• Domain & Range of Inverse Trigonometric Functions

The chapter explains principal value branches of inverse trigonometric functions.

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Important Formulas

Principal Value Range of sin⁻¹x

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

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Principal Value Range of cos⁻¹x

$\cos^{-1}x\in[0,\pi]$

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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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Principal Value Range of cot⁻¹x

$\cot^{-1}x\in(0,\pi)$

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Questions & Step-by-step Solutions

Question 1

Find:

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

Solution

We know:

sin(−π/6)=−1/2

Principal value range of:

sin⁻¹x

is:

[−π/2, π/2]

Therefore:

Answer = −π/6

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Question 2

Find:

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

Solution

We know:

cos(π/6)=√3/2

Principal value range of:

cos⁻¹x

is:

[0, π]

Therefore:

Answer = π/6

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Question 3

Find:

$\cosec^{-1}(2)$

Solution

We know:

cosec(π/6)=2

Therefore:

Answer = π/6

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Question 4

Find:

$\tan^{-1}(-\sqrt{3})$

Solution

We know:

tan(−π/3)=−√3

Therefore:

Answer = −π/3

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Question 5

Find:

$\cos^{-1}\left(-\frac{1}{2}\right)$

Solution

We know:

cos(2π/3)=−1/2

Since principal value range of cosine inverse is:

[0, π]

Therefore:

Answer = 2π/3

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Question 6

Find:

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

Solution

We know:

tan(−π/4)=−1

Therefore:

Answer = −π/4

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Question 7

Find:

$\sec^{-1}\left(-\frac{2}{\sqrt{3}}\right)$

Solution

We know:

sec(5π/6)=−2/√3

Therefore:

Answer = 5π/6

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Question 8

Find:

$\cot^{-1}(\sqrt{3})$

Solution

We know:

cot(π/6)=√3

Therefore:

Answer = π/6

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Question 9

Find:

$\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)$

Solution

We know:

cos(3π/4)=−1/√2

Therefore:

Answer = 3π/4

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Question 10

Find:

$\cosec^{-1}(-\sqrt{2})$

Solution

We know:

cosec(−π/4)=−√2

Therefore:

Answer = −π/4

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Question 11

Find the value of:

$\tan^{-1}(1)+\cos^{-1}\left(-\frac{1}{2}\right)+\sin^{-1}\left(-\frac{1}{2}\right)$

Solution

Using standard values:

tan⁻¹(1)=π/4

cos⁻¹(−1/2)=2π/3

sin⁻¹(−1/2)=−π/6

Adding:

π/4 + 2π/3 − π/6

LCM = 12

= 3π/12 + 8π/12 − 2π/12

Therefore:

Answer = 3π/4

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Question 12

Find:

$\cos^{-1}\left(\frac{1}{2}\right)+2\sin^{-1}\left(\frac{1}{2}\right)$

Solution

Using standard values:

cos⁻¹(1/2)=π/3

sin⁻¹(1/2)=π/6

Hence:

π/3 + 2(π/6)

= π/3 + π/3

Therefore:

Answer = 2π/3

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Question 13

If:

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

then:

Solution

Range of:

sin⁻¹x

is:

[−π/2, π/2]

Therefore correct option is:

(B)

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Question 14

Evaluate:

$\tan^{-1}(\sqrt{3})-\sec^{-1}(-2)$

Solution

We know:

tan⁻¹(√3)=π/3

and:

sec⁻¹(−2)=2π/3

Therefore:

π/3 − 2π/3

Hence:

Answer = −π/3

Correct option:

(B)

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Common Mistakes

• Ignoring principal value ranges
• Confusing sin⁻¹x with 1/sinx
• Using wrong quadrant values
• Forgetting domain restrictions
• Wrong sign in inverse functions

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Exam Tips

• Memorize standard inverse trigonometric values.
• Always check principal value branch.
• Use unit circle for quick answers.
• Practice simplification regularly.

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Practice MCQs

MCQ 1

Principal value range of:

$\cos^{-1}x$

is:

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

Answer:

B. [0, π]

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MCQ 2

Value of:

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

is:

A. π
B. π/2
C. π/4
D. 0

Answer:

C. π/4

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MCQ 3

Value of:

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

is:

A. π/6
B. −π/6
C. 5π/6
D. −5π/6

Answer:

B. −π/6

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FAQ Section

What is principal value?

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

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What is the range of sin⁻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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What is the range of tan⁻¹x?

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

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