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CBSE · Class 12 · Mathematics

Inverse Trigonometric Functions

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
The principal value of cos−1(1/√2) is:
  1. π/4
  2. −π/4
  3. 3π/4
  4. π/6
2.
tan−1(tan(4π/3)) equals:
  1. π/3
  2. 4π/3
  3. −π/3
  4. 2π/3
3.
cot(cos−1(5/13)) equals:
  1. 5/12
  2. 12/5
  3. 12/13
  4. 5/13
4.
If cosec−1 x = 2π/7, then sec−1 x equals:
  1. −2π/7
  2. 2π/7
  3. 5π/7
  4. 3π/14
5.
The principal value branch of sin−1 is:
  1. (0, π)
  2. [−π/2, π/2]
  3. [0, π]
  4. (−π/2, π/2)
6.
tan−1 (5/6) + tan−1 (1/7) equals:
  1. tan−1 (41/37)
  2. tan−1 (29/47)
  3. tan−1 (41/42)
  4. tan−1 (41/47)
7.
If sin−1 x = 7π/18, then sin−1(−x) equals:
  1. π/9
  2. 11π/18
  3. 7π/18
  4. −7π/18
8.
Assertion (A): cos−1(cos(π/2)) = π/2.
Reason (R): cos−1(cos θ) = θ only when θ lies in the principal branch of cos−1.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
Find the principal values of (a) sec−1(2/√3), (b) sin−1(1), (c) sin−1(−1/√2).
10.
Find the value of cos−1(1/2) + sec−1(−2) − sec−1(−1).
11.
Find the value of tan−1(tan(4π/3)).
12.
Find the value of tan(cos−1(12/13)).
13.
Find the value of cos(cos−1(√3/2) + cosec−1(−2/√3)).
14.
Solve cot−1(4x + 5) = π/4.
15.
Show that tan−1 (5/6) + tan−1 (1/11) = π/4.
16.
Find the domain of f(x) = sin−1(3x − 4).
Section C — Word problems4 × 4 = 16 marks
17.
Find the principal values of cosec−1(2/√3), cot−1(√3), sec−1(−1), and hence the value of their sum.
18.
Write tan−1 (4/7) + tan−1 (1/3) as a single inverse tangent, and say why the formula applies.
19.
Let θ = cos−1(9/41). Find sin θ, cos θ, tan θ, sin 2θ and cos 2θ.
20.
Solve tan−1 6x + tan−1 20x = π/4.
AmbifulWorksheets
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CBSE · Class 12 · Mathematics

Answer key

Inverse Trigonometric Functions · AW-12M-ITF-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(a) π/4the value must lie in the principal branch of cos−1
2(a) π/34π/3 is outside the principal branch; tan(4π/3) = tan(π/3) and π/3 lies in it
3(a) 5/12a right triangle with opposite side 12, adjacent side 5, hypotenuse 13
4(d) 3π/14cosec−1 x + sec−1 x = π/2
5(b) [−π/2, π/2]—
6(a) tan−1 (41/37)tan−1 a + tan−1 b = tan−1((a + b)/(1 − ab)) for ab < 1: (41/42)/(37/42)
7(d) −7π/18sin−1(−x) = −sin−1 x
8(a) Both A and R are true, and R is the correct explanation of A.A is true: π/2 lies in the principal branch of cos−1. R is true and explains A.
9(a) π/6; (b) π/2; (c) −π/4pick the angle in each principal branch
100cos−1(1/2) = π/3, sec−1(−2) = 2π/3, sec−1(−1) = π
11π/3tan(4π/3) = tan(π/3), and π/3 is in the principal branch
125/12let θ = cos−1(12/13); the triangle has sides 5, 12, 13
13√3/2π/6 − π/3 = −π/6
14x = −14x + 5 = cot(π/4) = 1
15π/4(a + b)/(1 − ab) = (61/66)/(61/66) = 1, and ab < 1
16[1, 5/3]−1 ≤ 3x − 4 ≤ 1
17cosec−1(2/√3) = π/3, cot−1(√3) = π/6, sec−1(−1) = π; sum = 3π/2each value from its principal branch
18tan−1 (19/17)ab = 4/21 < 1, so tan−1 a + tan−1 b = tan−1((a + b)/(1 − ab)) = tan−1((19/21)/(17/21))
19sin θ = 40/41, cos θ = 9/41, tan θ = 40/9, sin 2θ = 720/1681, cos 2θ = −1519/1681sin 2θ = 2 sin θ cos θ; cos 2θ = cos2θ − sin2θ
20x = 1/30(26x)/(1 − 120x2) = 1 ⇒ 120x2 + 26x − 1 = 0 ⇒ x = 1/30 or x = −1/4; the negative root makes the left side negative, so reject it