NCERT Solutions Class 12 Maths Chapter 2 – Inverse Trigonometric Functions

NCERT Solutions for Class 12 Maths Chapter 2: Inverse Trigonometric Functions

This chapter covers the inverse of trigonometric functions – their domain, range, principal values, and important properties. These concepts are essential for calculus and are frequently asked in competitive exams.

Exercise 2.1 Solutions

Q1. Find the principal value of sin⁻¹(−1/2).

Solution:

Let sin⁻¹(−1/2) = y

sin y = −1/2

Since principal value of sin⁻¹ lies in [−π/2, π/2]

sin(−π/6) = −1/2

Therefore, y = −π/6

Q2. Find the principal value of cos⁻¹(√3/2).

Solution:

Let cos⁻¹(√3/2) = y

cos y = √3/2

Since principal value of cos⁻¹ lies in [0, π]

cos(π/6) = √3/2

Therefore, y = π/6

Q3. Find the principal value of tan⁻¹(−√3).

Solution:

Let tan⁻¹(−√3) = y

tan y = −√3

Since principal value of tan⁻¹ lies in (−π/2, π/2)

tan(−π/3) = −√3

Therefore, y = −π/3

Exercise 2.2 Solutions

Q1. Prove that sin⁻¹(2x√(1−x²)) = 2sin⁻¹x, −1/√2 ≤ x ≤ 1/√2

Proof:

Let sin⁻¹x = θ, so x = sin θ

RHS = 2θ

LHS = sin⁻¹(2x√(1−x²))

= sin⁻¹(2sin θ √(1−sin²θ))

= sin⁻¹(2sin θ cos θ)

= sin⁻¹(sin 2θ)

= 2θ = RHS

Hence proved.

Q2. Prove that tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1−xy)), xy < 1

Proof:

Let tan⁻¹x = A and tan⁻¹y = B

Then tan A = x and tan B = y

tan(A + B) = (tan A + tan B)/(1 − tan A tan B)

= (x + y)/(1 − xy)

A + B = tan⁻¹((x + y)/(1 − xy))

tan⁻¹x + tan⁻¹y = tan⁻¹((x + y)/(1 − xy))

Hence proved.

Principal Value Ranges

Key Takeaways

• Inverse trig functions exist only in restricted domains
• sin⁻¹x + cos⁻¹x = π/2
• tan⁻¹x + cot⁻¹x = π/2
• sec⁻¹x + cosec⁻¹x = π/2
• 2tan⁻¹x = sin⁻¹(2x/(1+x²)) = cos⁻¹((1−x²)/(1+x²))
• 3tan⁻¹x = tan⁻¹((3x−x³)/(1−3x²))