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NCERT Solutions · Class 12 · Chapter 2: Inverse Trigonometric Functions

NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.1

Exercise 2.1: Principal values. The principal value is the one angle in the function's principal branch whose ratio is the given number. Principal value branches: \(\sin^{-1}\): \(\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]\), \(\cos^{-1}\): \([0, \pi]\), \(\tan^{-1}\): \(\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)\), \(\cot^{-1}\): \((0, \pi)\), \(\sec^{-1}\): \([0, \pi] - \left\{\tfrac{\pi}{2}\right\}\), \(\csc^{-1}\): \(\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right] - \{0\}\). Negative inputs: \(\sin^{-1}(-x) = -\sin^{-1}x\) but \(\cos^{-1}(-x) = \pi - \cos^{-1}x\).

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Exercise 2.1 questions and solutions

Exercise 2.1, Question 1

Find the principal value.
\(\sin^{-1}\left(-\tfrac12\right)\)
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  1. We need \[y \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]\] with \(\sin y = -\tfrac12\).
  2. \[\sin\left(-\tfrac{\pi}{6}\right) = -\tfrac12\]
Answer: \(-\tfrac{\pi}{6}\)

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Exercise 2.1, Question 2

\(\cos^{-1}\left(\tfrac{\sqrt3}{2}\right)\)
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  1. \(y \in [0, \pi]\) with \(\cos y = \tfrac{\sqrt3}{2}\): \(y = \tfrac{\pi}{6}\).
Answer: \(\tfrac{\pi}{6}\)

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Exercise 2.1, Question 3

\(\csc^{-1}(2)\)
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  1. \[\begin{aligned}\csc y &= 2 \Leftrightarrow \sin y \\ &= \tfrac12\end{aligned}\], with \[y \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right] - \{0\}\]: \(y = \tfrac{\pi}{6}\).
Answer: \(\tfrac{\pi}{6}\)

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Exercise 2.1, Question 4

\(\tan^{-1}\left(-\sqrt3\right)\)
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  1. \(\tan\tfrac{\pi}{3} = \sqrt3\) and tan is odd, so \[\tan\left(-\tfrac{\pi}{3}\right) = -\sqrt3\]; \[-\tfrac{\pi}{3} \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)\]
Answer: \(-\tfrac{\pi}{3}\)

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Exercise 2.1, Question 5

\(\cos^{-1}\left(-\tfrac12\right)\)
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  1. \(\cos^{-1}(-x) = \pi - \cos^{-1}x\).
  2. \(\pi - \tfrac{\pi}{3} = \tfrac{2\pi}{3}\), which lies in \([0, \pi]\).
Answer: \(\tfrac{2\pi}{3}\)

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Exercise 2.1, Question 6

\(\tan^{-1}(-1)\)
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  1. \(\tan\left(-\tfrac{\pi}{4}\right) = -1\) and \[-\tfrac{\pi}{4} \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)\]
Answer: \(-\tfrac{\pi}{4}\)

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Exercise 2.1, Question 7

\(\sec^{-1}\left(\tfrac{2}{\sqrt3}\right)\)
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  1. \[\begin{aligned}\sec y &= \tfrac{2}{\sqrt3} \Leftrightarrow \cos y \\ &= \tfrac{\sqrt3}{2}\end{aligned}\], \(y \in [0, \pi]\), \(y \ne \tfrac{\pi}{2}\): \(y = \tfrac{\pi}{6}\).
Answer: \(\tfrac{\pi}{6}\)

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Exercise 2.1, Question 10

\(\csc^{-1}\left(-\sqrt2\right)\)
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  1. \[\begin{aligned}\csc y &= -\sqrt2 \Leftrightarrow \sin y \\ &= -\tfrac{1}{\sqrt2}\end{aligned}\], \[y \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right] - \{0\}\]: \(y = -\tfrac{\pi}{4}\).
Answer: \(-\tfrac{\pi}{4}\)

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Exercise 2.1, Question 11

Find the value.
\(\tan^{-1}(1) + \cos^{-1}\left(-\tfrac12\right) + \sin^{-1}\left(-\tfrac12\right)\)
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  1. \(\tan^{-1}1 = \tfrac{\pi}{4}\), \[\cos^{-1}\left(-\tfrac12\right) = \tfrac{2\pi}{3}\], \[\sin^{-1}\left(-\tfrac12\right) = -\tfrac{\pi}{6}\]
  2. \[\begin{aligned}\tfrac{\pi}{4} + \tfrac{2\pi}{3} - \tfrac{\pi}{6} &= \tfrac{3\pi + 8\pi - 2\pi}{12} \\ &= \tfrac{9\pi}{12}\end{aligned}\]
Answer: \(\tfrac{3\pi}{4}\)

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Exercise 2.1, Question 12

\(\cos^{-1}\left(\tfrac12\right) + 2\sin^{-1}\left(\tfrac12\right)\)
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  1. \(\tfrac{\pi}{3} + 2 \cdot \tfrac{\pi}{6}\).
Answer: \(\tfrac{2\pi}{3}\)

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Exercise 2.1, Question 13

If \(\sin^{-1}x = y\), then: (A) \(0 \le y \le \pi\) (B) \(-\tfrac{\pi}{2} \le y \le \tfrac{\pi}{2}\) (C) \(0 < y < \pi\) (D) \(-\tfrac{\pi}{2} < y < \tfrac{\pi}{2}\)
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  1. The principal branch of \(\sin^{-1}\) is the closed interval \[\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]\] (the ends are reached at \(x = \pm1\)).
Answer: (B)

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Exercise 2.1, Question 14

\(\tan^{-1}\sqrt3 - \sec^{-1}(-2)\) equals: (A) \(\pi\) (B) \(-\tfrac{\pi}{3}\) (C) \(\tfrac{\pi}{3}\) (D) \(\tfrac{2\pi}{3}\)
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  1. \(\tan^{-1}\sqrt3 = \tfrac{\pi}{3}\); \[\begin{aligned}\sec^{-1}(-2) &= \cos^{-1}\left(-\tfrac12\right) \\ &= \tfrac{2\pi}{3}\end{aligned}\]
  2. \[\tfrac{\pi}{3} - \tfrac{2\pi}{3} = -\tfrac{\pi}{3}\]
Answer: (B) \(-\tfrac{\pi}{3}\)

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Inverse Trigonometric Functions in our sample papers: Sample paper 1 (questions 2, 21) · Sample paper 2 (questions 2, 21) · Sample paper 3 (questions 2, 21) · Sample paper 4 (questions 2, 21) · Sample paper 5 (questions 2, 21).

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Textbook: NCERT Mathematics Class 12, Parts I and II (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in. Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log.