CBSE Class 12 · Chapter 2 · Relations and Functions · 2026-27

Inverse Trigonometric Functions Class 12: MCQ and case study questions

15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.

15 MCQs (1 mark each)
2 case studies (4 marks each)
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Multiple-choice questions

Tap an option to check it; the answer and a worked solution open. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.

Q1

·1 mark·Multiple choicePrincipal values

Of the angles below, which one is the principal value of $\sin^{- 1} (- \frac{1}{2})$ ?

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Answer: (b)

- \frac{π}{6}

Why: The principal value lies in [−π/2, π/2] and sin(−π/6) = −1/2.

We need $θ \in [- \frac{π}{2}, \frac{π}{2}]$ with $\sin θ = - \frac{1}{2}$ : $θ = - \frac{π}{6}$ . ( $\frac{7 π}{6}$ and $\frac{11 π}{6}$ have the right sine but are outside the branch.)

Q2

·1 mark·Multiple choicePrincipal values

The principal value of $\cos^{- 1} (- \frac{\sqrt{3}}{2})$ is

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Answer: (c)

\frac{5 π}{6}

Why: The principal value lies in [0, π] and cos(5π/6) = −√3/2.

$θ \in [0, π]$ with $\cos θ = - \frac{\sqrt{3}}{2}$ : $θ = π - \frac{π}{6} = \frac{5 π}{6}$ .

Q3

·1 mark·Multiple choicePrincipal values

$\tan^{- 1} (- \frac{1}{\sqrt{3}})$ , taking the principal value, equals

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Answer: (a)

- \frac{π}{6}

Why: The principal value lies in (−π/2, π/2) and tan(−π/6) = −1/√3.

$θ \in (- \frac{π}{2}, \frac{π}{2})$ with $\tan θ = - \frac{1}{\sqrt{3}}$ : $θ = - \frac{π}{6}$ . ( $\frac{5 π}{6}$ has the same tangent but is outside the branch.)

Q4

·1 mark·Multiple choiceDomain and range of principal value branches

Which of the following is not a value taken by the principal branch of $\sin^{- 1} x$ ?

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Answer: (d)

\frac{2 π}{3}

Why: The principal branch of sin⁻¹ takes values only in [−π/2, π/2].

The range of the principal branch of $\sin^{- 1}$ is $[- \frac{π}{2}, \frac{π}{2}]$ : $- \frac{π}{2} = \sin^{- 1} (- 1)$ , $0 = \sin^{- 1} 0$ , $\frac{π}{3} = \sin^{- 1} \frac{\sqrt{3}}{2}$ . $\frac{2 π}{3} > \frac{π}{2}$ is never taken.

Q5

·1 mark·Multiple choiceDomain and range of principal value branches

The domain of $f (x) = \sin^{- 1} (3 x + 2)$ is

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Answer: (a)

[- 1, - \frac{1}{3}]

Why: Need −1 ≤ 3x + 2 ≤ 1.

$- 1 \leq 3 x + 2 \leq 1 \Rightarrow - 3 \leq 3 x \leq - 1 \Rightarrow - 1 \leq x \leq - \frac{1}{3}$ .

Q6

·1 mark·Multiple choiceEvaluating expressions

The value of ${cosec}^{- 1} (2) + \sec^{- 1} (2)$ is

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Answer: (c)

\frac{π}{2}

Why: cosec⁻¹ 2 = π/6 and sec⁻¹ 2 = π/3 (principal values).

${cosec}^{- 1} 2 = \frac{π}{6}$ (as $\sin \frac{π}{6} = \frac{1}{2}$ ) and $\sec^{- 1} 2 = \frac{π}{3}$ (as $\cos \frac{π}{3} = \frac{1}{2}$ ). Sum $= \frac{π}{2}$ .

Q7

·1 mark·Multiple choiceEvaluating expressions

The value of $\tan^{- 1} (\tan \frac{3 π}{4})$ is

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Answer: (b)

- \frac{π}{4}

Why: tan(3π/4) = −1, and the principal value of tan⁻¹(−1) is −π/4.

$\frac{3 π}{4}$ is outside $(- \frac{π}{2}, \frac{π}{2})$ . $\tan \frac{3 π}{4} = - 1$ , so the value is $\tan^{- 1} (- 1) = - \frac{π}{4}$ .

Q8

·1 mark·Multiple choiceEvaluating expressions

The value of $\cot (\sin^{- 1} \frac{8}{17})$ is

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Answer: (d)

\frac{15}{8}

Why: If sin θ = 8/17 with θ acute, cos θ = 15/17, so cot θ = 15/8.

Let $θ = \sin^{- 1} \frac{8}{17} \in (0, \frac{π}{2})$ . Then $\cos θ = \sqrt{1 - \frac{64}{289}} = \frac{15}{17}$ and $\cot θ = \frac{15}{8}$ .

Q9

·1 mark·Multiple choiceEvaluating expressions

The value of $\sec^{- 1} (\sqrt{2}) + \cot^{- 1} (\sqrt{3})$ is

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Answer: (a)

\frac{5 π}{12}

Why: sec⁻¹ √2 = π/4 and cot⁻¹ √3 = π/6.

$\sec^{- 1} \sqrt{2} = \frac{π}{4}$ ( $\cos \frac{π}{4} = \frac{1}{\sqrt{2}}$ ); $\cot^{- 1} \sqrt{3} = \frac{π}{6}$ . Sum $= \frac{3 π + 2 π}{12} = \frac{5 π}{12}$ .

Q10

·1 mark·Multiple choicePrincipal values

If $\cos^{- 1} x = \frac{2 π}{3}$ , then $x$ equals

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Answer: (b)

- \frac{1}{2}

Why: x = cos(2π/3), and 2π/3 is in [0, π].

$\frac{2 π}{3} \in [0, π]$ , so $x = \cos \frac{2 π}{3} = - \frac{1}{2}$ .

Q11

·1 mark·Multiple choiceEvaluating expressions

The value of $\cos (\frac{π}{6} + \cos^{- 1} (- \frac{\sqrt{3}}{2}))$ is

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Answer: (c)

- 1

Why: cos⁻¹(−√3/2) = 5π/6, so the angle is π.

$\cos^{- 1} (- \frac{\sqrt{3}}{2}) = \frac{5 π}{6}$ . $\cos (\frac{π}{6} + \frac{5 π}{6}) = \cos π = - 1$ .

Q12

·1 mark·Multiple choiceEvaluating expressions

The value of $\tan (2 \tan^{- 1} \frac{1}{2})$ is

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Answer: (c)

\frac{4}{3}

Why: With tan θ = 1/2, tan 2θ = 2 tan θ/(1 − tan² θ).

Let $θ = \tan^{- 1} \frac{1}{2}$ . $\tan 2 θ = \frac{2 \cdot \frac{1}{2}}{1 - \frac{1}{4}} = \frac{1}{3 / 4} = \frac{4}{3}$ .

Q13

·1 mark·Multiple choiceEvaluating expressions

The value of $\sin^{- 1} (\cos \frac{π}{3})$ is

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Answer: (d)

\frac{π}{6}

Why: cos(π/3) = 1/2 and sin⁻¹(1/2) = π/6.

$\cos \frac{π}{3} = \frac{1}{2}$ , and the principal value of $\sin^{- 1} \frac{1}{2}$ is $\frac{π}{6}$ .

Q14

·1 mark·Multiple choiceDomain and range of principal value branches

The domain of $f (x) = \cos^{- 1} (x^{2})$ is

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Answer: (d)

[- 1, 1]

Why: Need −1 ≤ x² ≤ 1, i.e. x² ≤ 1.

$x^{2} \geq 0 \geq - 1$ always, and $x^{2} \leq 1 \Leftrightarrow - 1 \leq x \leq 1$ . Domain $= [- 1, 1]$ .

Q15

·1 mark·Multiple choicePrincipal values

Let $θ$ be the principal value of $\cot^{- 1} (- \sqrt{3})$ , which lies in $(0, π)$ . Then $\sin θ$ equals

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Answer: (b)

\frac{1}{2}

Why: θ = 5π/6 (cot 5π/6 = −√3), and sin 5π/6 = 1/2.

In $(0, π)$ , $\cot θ = - \sqrt{3}$ gives $θ = π - \frac{π}{6} = \frac{5 π}{6}$ . So $\sin θ = \sin \frac{5 π}{6} = \frac{1}{2}$ (positive, as $θ$ is in $(0, π)$ ).

Case study questions

Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part. All Class 12 case study questions by chapter.

Case study 1: Ramps at the community centre (4 marks)

A community centre builds two straight ramps. Ramp A rises $1$ m over a horizontal run of $\sqrt{3}$ m; ramp B, for loading goods, rises $\sqrt{3}$ m over a horizontal run of $1$ m. The angle a ramp makes with the ground is $θ = \tan^{- 1} (\frac{rise}{run})$ .

(i) Find the angle of ramp A in radians. [1 mark]

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Answer:

\frac{π}{6}

θ_{A} = \tan^{- 1} \frac{1}{\sqrt{3}} = \frac{π}{6}

. A1

(ii) Find the angle of ramp B in radians. [1 mark]

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Answer:

\frac{π}{3}

θ_{B} = \tan^{- 1} \sqrt{3} = \frac{π}{3}

. A1

(iii) Find the length of ramp A and hence show that

θ_{A} = \sin^{- 1} \frac{1}{2}

. [2 marks]

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Answer:

2

Length

= \sqrt{1 + 3} = 2

m M1, so

\sin θ_{A} = \frac{1}{2}

and

θ_{A} = \sin^{- 1} \frac{1}{2} = \frac{π}{6}

. A1

OR Find

θ_{B} - θ_{A}

and the value of

\cos (θ_{B} - θ_{A})

. [2 marks]

Show answer

Answer:

\frac{π}{6}

;

\frac{\sqrt{3}}{2}

θ_{B} - θ_{A} = \frac{π}{3} - \frac{π}{6} = \frac{π}{6}

M1;

\cos \frac{π}{6} = \frac{\sqrt{3}}{2}

. A1

Case study 2: Playground swing (4 marks)

A swing seat hangs from a horizontal bar on chains $2$ m long. When the seat is pulled sideways so that it is $d$ m (horizontally) from its lowest position, with the chains kept straight, the chains make an angle $θ = \sin^{- 1} (\frac{d}{2})$ with the vertical.

(i) Find

θ

when

d = 1

. [1 mark]

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Answer:

\frac{π}{6}

θ = \sin^{- 1} \frac{1}{2} = \frac{π}{6}

. A1

(ii) Find

θ

when

d = \sqrt{2}

. [1 mark]

Show answer

Answer:

\frac{π}{4}

θ = \sin^{- 1} \frac{\sqrt{2}}{2} = \sin^{- 1} \frac{1}{\sqrt{2}} = \frac{π}{4}

. A1

(iii) Since

d

is a distance,

d \geq 0

. For which values of

d

θ = \sin^{- 1} (\frac{d}{2})

defined, and what is the largest value

θ

can take? [2 marks]

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Answer:

0 \leq d \leq 2

;

\frac{π}{2}

Need

- 1 \leq \frac{d}{2} \leq 1

, i.e.

- 2 \leq d \leq 2

; with

d \geq 0

this gives

0 \leq d \leq 2

M1; the largest principal value is

\sin^{- 1} 1 = \frac{π}{2}

(chains horizontal). A1

OR When

d = \sqrt{3}

, find

θ

and how high the seat has risen above its lowest position. [2 marks]

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Answer:

\frac{π}{3}

;

1

θ = \sin^{- 1} \frac{\sqrt{3}}{2} = \frac{π}{3}

M1; rise

= 2 - 2 \cos \frac{π}{3} = 2 - 1 = 1

m. A1

Next steps for Inverse Trigonometric Functions

This free set is separate from the chapter's question bank. On the Inverse Trigonometric Functions chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Pro adds all 47 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders, the full common-mistakes library and AI marking of your written working.

Inverse Trigonometric Functions: important questions What Pro includes

Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was checked twice, the second time by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.