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
Choose one option. 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 \(\sin^{-1}\left(-\dfrac12\right)\)?
(a)\(\dfrac{\pi}{6}\)
(b)\(-\dfrac{\pi}{6}\)
(c)\(\dfrac{7\pi}{6}\)
(d)\(\dfrac{11\pi}{6}\)
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Answer: (b) \(-\dfrac{\pi}{6}\)
Why: The principal value lies in [−π/2, π/2] and sin(−π/6) = −1/2.
We need \(\theta \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]\) with \(\sin\theta = -\tfrac12\): \(\theta = -\dfrac{\pi}{6}\). (\(\tfrac{7\pi}{6}\) and \(\tfrac{11\pi}{6}\) have the right sine but are outside the branch.)
Q2
·1 mark·Multiple choicePrincipal values
The principal value of \(\cos^{-1}\left(-\dfrac{\sqrt3}{2}\right)\) is
(a)\(\dfrac{\pi}{6}\)
(b)\(-\dfrac{\pi}{6}\)
(c)\(\dfrac{5\pi}{6}\)
(d)\(\dfrac{7\pi}{6}\)
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Answer: (c) \(\dfrac{5\pi}{6}\)
Why: The principal value lies in [0, π] and cos(5π/6) = −√3/2.
\(\tan^{-1}\left(-\dfrac{1}{\sqrt3}\right)\), taking the principal value, equals
(a)\(-\dfrac{\pi}{6}\)
(b)\(\dfrac{5\pi}{6}\)
(c)\(\dfrac{\pi}{6}\)
(d)\(-\dfrac{\pi}{3}\)
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Answer: (a) \(-\dfrac{\pi}{6}\)
Why: The principal value lies in (−π/2, π/2) and tan(−π/6) = −1/√3.
\(\theta \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)\) with \(\tan\theta = -\tfrac{1}{\sqrt3}\): \(\theta = -\dfrac{\pi}{6}\). (\(\tfrac{5\pi}{6}\) 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 \(\sin^{-1}x\)?
(a)\(-\dfrac{\pi}{2}\)
(b)\(0\)
(c)\(\dfrac{\pi}{3}\)
(d)\(\dfrac{2\pi}{3}\)
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Answer: (d) \(\dfrac{2\pi}{3}\)
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 \(\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]\): \(-\tfrac{\pi}{2} = \sin^{-1}(-1)\), \(0 = \sin^{-1}0\), \(\tfrac{\pi}{3} = \sin^{-1}\tfrac{\sqrt3}{2}\). \(\dfrac{2\pi}{3} \gt \dfrac{\pi}{2}\) is never taken.
Q5
·1 mark·Multiple choiceDomain and range of principal value branches
\(\operatorname{cosec}^{-1} 2 = \dfrac{\pi}{6}\) (as \(\sin\tfrac{\pi}{6} = \tfrac12\)) and \(\sec^{-1} 2 = \dfrac{\pi}{3}\) (as \(\cos\tfrac{\pi}{3} = \tfrac12\)). Sum \(= \dfrac{\pi}{2}\).
Q7
·1 mark·Multiple choiceEvaluating expressions
The value of \(\tan^{-1}\left(\tan\dfrac{3\pi}{4}\right)\) is
(a)\(\dfrac{3\pi}{4}\)
(b)\(-\dfrac{\pi}{4}\)
(c)\(\dfrac{\pi}{4}\)
(d)\(-\dfrac{3\pi}{4}\)
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Answer: (b) \(-\dfrac{\pi}{4}\)
Why: tan(3π/4) = −1, and the principal value of tan⁻¹(−1) is −π/4.
\(\dfrac{3\pi}{4}\) is outside \(\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)\). \(\tan\dfrac{3\pi}{4} = -1\), so the value is \(\tan^{-1}(-1) = -\dfrac{\pi}{4}\).
Q8
·1 mark·Multiple choiceEvaluating expressions
The value of \(\cot\left(\sin^{-1}\dfrac{8}{17}\right)\) is
(a)\(\dfrac{8}{15}\)
(b)\(\dfrac{17}{15}\)
(c)\(\dfrac{15}{17}\)
(d)\(\dfrac{15}{8}\)
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Answer: (d) \(\dfrac{15}{8}\)
Why: If sin θ = 8/17 with θ acute, cos θ = 15/17, so cot θ = 15/8.
Let \(\theta = \sin^{-1}\tfrac{8}{17} \in \left(0, \tfrac{\pi}{2}\right)\). Then \(\cos\theta = \sqrt{1 - \tfrac{64}{289}} = \tfrac{15}{17}\) and \(\cot\theta = \dfrac{15}{8}\).
Q9
·1 mark·Multiple choiceEvaluating expressions
The value of \(\sec^{-1}(\sqrt2) + \cot^{-1}(\sqrt3)\) is
The value of \(\sin^{-1}\left(\cos\dfrac{\pi}{3}\right)\) is
(a)\(\dfrac{\pi}{3}\)
(b)\(\dfrac{2\pi}{3}\)
(c)\(\dfrac{5\pi}{6}\)
(d)\(\dfrac{\pi}{6}\)
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Answer: (d) \(\dfrac{\pi}{6}\)
Why: cos(π/3) = 1/2 and sin⁻¹(1/2) = π/6.
\(\cos\dfrac{\pi}{3} = \dfrac12\), and the principal value of \(\sin^{-1}\tfrac12\) is \(\dfrac{\pi}{6}\).
Q14
·1 mark·Multiple choiceDomain and range of principal value branches
The domain of \(f(x) = \cos^{-1}(x^2)\) is
(a)\([0, 1]\)
(b)\([-1, 0]\)
(c)\(\mathbb{R}\)
(d)\([-1, 1]\)
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Answer: (d) \([-1, 1]\)
Why: Need −1 ≤ x² ≤ 1, i.e. x² ≤ 1.
\(x^2 \ge 0 \ge -1\) always, and \(x^2 \le 1 \Leftrightarrow -1 \le x \le 1\). Domain \(= [-1, 1]\).
Q15
·1 mark·Multiple choicePrincipal values
Let \(\theta\) be the principal value of \(\cot^{-1}(-\sqrt3)\), which lies in \((0, \pi)\). Then \(\sin\theta\) equals
(a)\(-\dfrac12\)
(b)\(\dfrac12\)
(c)\(\dfrac{\sqrt3}{2}\)
(d)\(-\dfrac{\sqrt3}{2}\)
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Answer: (b) \(\dfrac12\)
Why: θ = 5π/6 (cot 5π/6 = −√3), and sin 5π/6 = 1/2.
In \((0, \pi)\), \(\cot\theta = -\sqrt3\) gives \(\theta = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}\). So \(\sin\theta = \sin\dfrac{5\pi}{6} = \dfrac12\) (positive, as \(\theta\) is in \((0, \pi)\)).
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.
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 \(\sqrt3\) m; ramp B, for loading goods, rises \(\sqrt3\) m over a horizontal run of \(1\) m. The angle a ramp makes with the ground is \(\theta = \tan^{-1}\left(\dfrac{\text{rise}}{\text{run}}\right)\).
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 \(\theta = \sin^{-1}\left(\dfrac{d}{2}\right)\) with the vertical.
(iii) Since \(d\) is a distance, \(d \ge 0\). For which values of \(d\) is \(\theta = \sin^{-1}\left(\dfrac{d}{2}\right)\) defined, and what is the largest value \(\theta\) can take? [2 marks]
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Answer: \(0 \le d \le 2\); \(\dfrac{\pi}{2}\)
Need \(-1 \le \dfrac{d}{2} \le 1\), i.e. \(-2 \le d \le 2\); with \(d \ge 0\) this gives \(0 \le d \le 2\) M1; the largest principal value is \(\sin^{-1}1 = \dfrac{\pi}{2}\) (chains horizontal). A1
OR When \(d = \sqrt3\), find \(\theta\) and how high the seat has risen above its lowest position. [2 marks]
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 Essentials adds all 39 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 and the full common-mistakes library. There is no AI marking on CBSE Math Revision: you check your work against the marking scheme.
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 re-checked by computer algebra and 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.