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Inverse Trigonometric Functions – Class 12 Mathematics notes, important questions and MCQs

NCERT Class 12 Mathematics, chapter 2. Notes, key terms, important questions with answers and MCQs.

Chapter overview

Inverse trigonometric functions give angles from ratios. Each is defined on a restricted principal-value branch so that it is a function.

Principal value branches

sin⁻¹x: domain [−1, 1], range [−π/2, π/2]; cos⁻¹x: [−1, 1] → [0, π]; tan⁻¹x: ℝ → (−π/2, π/2).
sin⁻¹(−x) = −sin⁻¹x; cos⁻¹(−x) = π − cos⁻¹x; tan⁻¹(−x) = −tan⁻¹x.

Identities

sin⁻¹x + cos⁻¹x = π/2; tan⁻¹x + cot⁻¹x = π/2.
tan⁻¹x + tan⁻¹y = tan⁻¹[(x + y)/(1 − xy)] for xy < 1.
2tan⁻¹x = tan⁻¹[2x/(1 − x²)] for |x| < 1.

Key terms

Principal value: the value of an inverse trig function within its principal branch
Principal value branch: the restricted range on which an inverse trig function is defined, e.g. [−π/2, π/2] for sin⁻¹
Domain of sin⁻¹ and cos⁻¹: [−1, 1], because sine and cosine values lie only in this interval

Formulas

sin⁻¹x + cos⁻¹x = π/2
tan⁻¹x + tan⁻¹y = tan⁻¹[(x + y)/(1 − xy)]

Important questions with answers

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

cos⁻¹(−x) = π − cos⁻¹x, so π − π/4 = 3π/4.

Prove that 2 tan⁻¹(1/2) + tan⁻¹(1/7) = tan⁻¹(31/17).3 marks

2 tan⁻¹(1/2) = tan⁻¹[(2 × 1/2)/(1 − 1/4)] = tan⁻¹(4/3). Then tan⁻¹(4/3) + tan⁻¹(1/7) = tan⁻¹[(4/3 + 1/7)/(1 − 4/21)] = tan⁻¹[(31/21)/(17/21)] = tan⁻¹(31/17).

Write tan⁻¹[√((1 − cos x)/(1 + cos x))], 0 < x < π, in the simplest form, and solve tan⁻¹(2x) + tan⁻¹(3x) = π/4.5 marks

(1 − cos x)/(1 + cos x) = tan²(x/2), and for 0 < x < π, tan(x/2) > 0 with x/2 in (0, π/2), so the expression = x/2. For the equation: tan⁻¹[(5x)/(1 − 6x²)] = π/4, so 5x = 1 − 6x², 6x² + 5x − 1 = 0, (6x − 1)(x + 1) = 0, x = 1/6 or −1. x = −1 makes both terms negative (sum cannot be π/4), so x = 1/6.

Quick revision

Range of sin⁻¹ is [−π/2, π/2].
Range of cos⁻¹ is [0, π].
sin⁻¹x + cos⁻¹x = π/2.

Exam tips

Always check the answer lies in the principal range (e.g. cos⁻¹(cos 4π/3) = 2π/3, not 4π/3).
Convert to a single function first: cot⁻¹x = tan⁻¹(1/x) for x > 0, sec⁻¹x = cos⁻¹(1/x).

Practice MCQs

Principal value of sin⁻¹(1/2) is
1. π/6
2. π/3
3. π/4
4. 5π/6
Show answer
Answer: (a) π/6. sin π/6 = 1/2.
Range of cos⁻¹x is
1. [−π/2, π/2]
2. [0, π]
3. (0, π)
4. ℝ
Show answer
Answer: (b) [0, π]. Principal branch.
cos⁻¹(−1/2) equals
1. π/3
2. −π/3
3. 2π/3
4. 5π/6
Show answer
Answer: (c) 2π/3. π − π/3.
sin⁻¹(0.3) + cos⁻¹(0.3) equals
1. 0
2. π/2
3. π
4. 0.6
Show answer
Answer: (b) π/2. Identity.
tan⁻¹(√3) − sec⁻¹(−2) equals
1. −π/3
2. π/3
3. π
4. 2π/3
Show answer
Answer: (a) −π/3. π/3 − 2π/3.
tan⁻¹(1/2) + tan⁻¹(1/3) equals
1. π/6
2. π/4
3. π/3
4. π/2
Show answer
Answer: (b) π/4. (5/6)/(1 − 1/6) = 1.
Principal value of cos⁻¹(cos 7π/6) is
1. 7π/6
2. −π/6
3. π/6
4. 5π/6
Show answer
Answer: (d) 5π/6. Must lie in [0, π].
sin(cos⁻¹(3/5)) equals
1. 3/5
2. 4/5
3. 5/4
4. 3/4
Show answer
Answer: (b) 4/5. Right triangle 3-4-5.

All chapters of Class 12 Mathematics

Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.