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Class 12 Mathematics

Chapter 2 Solutions — Inverse Trigonometric Functions

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Solutions

Overview

Step-by-step NCERT solutions for Inverse Trigonometric Functions (Chapter 2, CBSE Class 12 Mathematics) — every question and answer worked out in full, not just the final result. You can also read the Inverse Trigonometric Functions textbook chapter.

Solved

What these solutions cover

All 43 questions in Inverse Trigonometric Functions are solved in the PDF. Here's what's inside, exercise by exercise:

Exercise 2.1

  1. Find the principal value of sin⁻¹(-1/2).
  2. Find the principal value of cos⁻¹(√3/2).
  3. Find the principal value of cosec⁻¹(2).
  4. Find the principal value of tan⁻¹(-√3).
  5. Find the principal value of cos⁻¹(-1/2).
  6. Find the principal value of tan⁻¹(-1).
  7. Find the principal value of sec⁻¹(2/√3).
  8. Find the principal value of cot⁻¹(√3).
  9. Find the principal value of cos⁻¹(-1/√2).
  10. Find the principal value of cosec⁻¹(-√2).
  11. Find the value of tan⁻¹(1) + cos⁻¹(-1/2) + sin⁻¹(-1/2).
  12. Find the value of cos⁻¹(1/2) + 2 sin⁻¹(1/2).
  13. If sin⁻¹ x = y, then (A) 0 ≤ y ≤ π (B) -π/2 ≤ y ≤ π/2 (C) 0 < y < π (D) -π/2 < y < π/2
  14. tan⁻¹(√3) - sec⁻¹(-2) is equal to (A) π (B) -π/3 (C) π/3 (D) 2π/3

Exercise 2.2

  1. Prove that 3 sin⁻¹ x = sin⁻¹(3x - 4x³), x ∈ [-1/2, 1/2].
  2. Prove that 3 cos⁻¹ x = cos⁻¹(4x³ - 3x), x ∈ [1/2, 1].
  3. Write tan⁻¹((√(1+x²) - 1)/x), x ≠ 0 in the simplest form.
  4. Write tan⁻¹(√((1 - cos x)/(1 + cos x))), 0 < x < π in the simplest form.
  5. Write tan⁻¹((cos x - sin x)/(cos x + sin x)), -π/4 < x < 3π/4 in the simplest form.
  6. Write tan⁻¹(x / √(a² - x²)), |x| < a in the simplest form.
  7. Write tan⁻¹((3a² x - x³)/(a³ - 3ax²)), a > 0; -a/√3 < x < a/√3 in the simplest form.
  8. Find the value of tan⁻¹[2 cos(2 sin⁻¹(1/2))].
  9. Find the value of tan[(1/2)(sin⁻¹(2x/(1+x²)) + cos⁻¹((1-y²)/(1+y²)))], |x| < 1, y > 0 and xy < 1.
  10. Find the value of sin⁻¹(sin(2π/3)).
  11. Find the value of tan⁻¹(tan(3π/4)).
  12. Find the value of tan(sin⁻¹(3/5) + cot⁻¹(3/2)).
  13. cos⁻¹(cos(7π/6)) is equal to (A) 7π/6 (B) 5π/6 (C) π/3 (D) π/6
  14. sin(π/3 - sin⁻¹(-1/2)) is equal to (A) 1/2 (B) 1/3 (C) 1/4 (D) 1
  15. tan⁻¹(√3) - cot⁻¹(-√3) is equal to (A) π (B) -π/2 (C) 0 (D) 2√3

Miscellaneous Exercise

  1. Find the value of cos⁻¹(cos(13π/6)).
  2. Find the value of tan⁻¹(tan(7π/6)).
  3. Prove that 2 sin⁻¹(3/5) = tan⁻¹(24/7).
  4. Prove that sin⁻¹(8/17) + sin⁻¹(3/5) = tan⁻¹(77/36).
  5. Prove that cos⁻¹(4/5) + cos⁻¹(12/13) = cos⁻¹(33/65).
  6. Prove that cos⁻¹(12/13) + sin⁻¹(3/5) = sin⁻¹(56/65).
  7. Prove that tan⁻¹(63/16) = sin⁻¹(5/13) + cos⁻¹(3/5).
  8. Prove that tan⁻¹(√x) = (1/2) cos⁻¹((1-x)/(1+x)), x ∈ [0, 1].
  9. Prove that cot⁻¹((√(1+sin x) + √(1-sin x))/(√(1+sin x) - √(1-sin x))) = x/2, x ∈ (0, π/4).
  10. Prove that tan⁻¹((√(1+x) - √(1-x))/(√(1+x) + √(1-x))) = π/4 - (1/2) cos⁻¹ x, -1/√2 ≤ x ≤ 1. [Hint: Put x = cos 2θ]
  11. Solve 2 tan⁻¹(cos x) = tan⁻¹(2 cosec x).
  12. Solve tan⁻¹((1-x)/(1+x)) = (1/2) tan⁻¹ x, (x > 0).
  13. sin(tan⁻¹ x), |x| < 1 is equal to (A) x/√(1-x²) (B) 1/√(1-x²) (C) 1/√(1+x²) (D) x/√(1+x²)
  14. sin⁻¹(1-x) - 2 sin⁻¹ x = π/2, then x is equal to (A) 0, 1/2 (B) 1, 1/2 (C) 0 (D) 1/2
Sample

Worked examples

3 questions from Inverse Trigonometric Functions, solved step by step.

Exercise 2.1

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

Let sin⁻¹(-1/2) = y. Then sin y = -1/2.

The range of the principal value branch of sin⁻¹ is [-π/2, π/2].

We know sin(-π/6) = -1/2 and -π/6 lies in [-π/2, π/2].

Therefore, the principal value of sin⁻¹(-1/2) is -π/6.

Answer: -π/6

Exercise 2.2

Prove that 3 sin⁻¹ x = sin⁻¹(3x - 4x³), x ∈ [-1/2, 1/2].

Let x = sin θ, so θ = sin⁻¹ x. For x ∈ [-1/2, 1/2], θ ∈ [-π/6, π/6], which gives 3θ ∈ [-π/2, π/2].

Consider sin⁻¹(3x - 4x³) = sin⁻¹(3 sin θ - 4 sin³ θ).

Using the triple angle formula: sin 3θ = 3 sin θ - 4 sin³ θ.

So sin⁻¹(3x - 4x³) = sin⁻¹(sin 3θ) = 3θ (since 3θ ∈ [-π/2, π/2]).

And 3θ = 3 sin⁻¹ x.

Therefore, 3 sin⁻¹ x = sin⁻¹(3x - 4x³). (Proved)

Answer: Proved using the substitution x = sin θ and the identity sin 3θ = 3 sin θ - 4 sin³ θ.

Miscellaneous Exercise

Find the value of cos⁻¹(cos(13π/6)).

Note that 13π/6 does not lie in [0, π], the principal value branch of cos⁻¹.

Reduce 13π/6 modulo 2π: 13π/6 - 2π = 13π/6 - 12π/6 = π/6.

So cos(13π/6) = cos(π/6) = √3/2.

Since π/6 ∈ [0, π]: cos⁻¹(cos(13π/6)) = cos⁻¹(cos(π/6)) = π/6.

Answer: π/6

All 43 questions in Inverse Trigonometric Functions are solved step by step in the Solutions PDF above.

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