Inverse Trigonometric Functions: CBSE Class 12 Previous Year Questions

25 different questions from Inverse Trigonometric Functions (NCERT Chapter 2) asked in CBSE Class 12 Maths board exams 2026. Pick a mark group to practise, each with answers and step-by-step solutions.

1 mark questions11 questions 2 marks questions11 questions 5 marks questions3 questions

Most asked Inverse Trigonometric Functions questions

Q31 markMCQInverse Trigonometric Functions CBSE 2026 · 65/4/1

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

(A) $- \frac{π}{2}$
(B) $- \frac{π}{4}$
(C) $\frac{π}{4}$
(D) $\frac{π}{2}$

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Answer: (B)

- \frac{π}{4}

$sec^{- 1} (2) = \frac{π}{4}$ (principal range $[0, π] - {\frac{π}{2}}$ ).
$cosec^{- 1} (- 2) = - \frac{π}{4}$ (principal range $[- \frac{π}{2}, \frac{π}{2}] - {0}$ ).
Value $= \frac{π}{4} + 2 (- \frac{π}{4}) = - \frac{π}{4}$ .

Also asked in: 2026 65/4/2, 2026 65/4/3

Q31 markMCQInverse Trigonometric Functions CBSE 2026 · 65/5/1

For the inverse trigonometric functions, which of the following Principal Value Branch is not correctly defined ?

(A) $tan^{- 1} : R \to (- \frac{π}{2}, \frac{π}{2})$
(B) $sec^{- 1} : R - (- 1, 1) \to [0, π] - {\frac{π}{2}}$
(C) $cot^{- 1} : R \to (0, π)$
(D) $cosec^{- 1} : R - (- 1, 1) \to [- \frac{π}{2}, \frac{π}{2}]$

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Answer: (D)

cosec^{- 1} : R - (- 1, 1) \to [- \frac{π}{2}, \frac{π}{2}]

The principal value branch of $cosec^{- 1}$ is $[- \frac{π}{2}, \frac{π}{2}] - {0}$ , since $cosec y$ is not defined at $y = 0$ .
Options (A), (B) and (C) are the standard principal value branches, so (D) is not correctly defined.

Also asked in: 2026 65/5/2, 2026 65/5/3

Q252 marksVery Short AnswerInverse Trigonometric Functions CBSE 2026 · 65/1/1

Simplify : $tan^{- 1} (\frac{c o s 2 x - s i n 2 x}{c o s 2 x + s i n 2 x}), 0 < x < \frac{π}{4}$ .

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Answer:

\frac{π}{4} - 2 x

Divide numerator and denominator by $cos 2 x$ : $\frac{1 - t a n 2 x}{1 + t a n 2 x} = tan (\frac{π}{4} - 2 x)$ .
For $0 < x < \frac{π}{4}$ , $- \frac{π}{4} < \frac{π}{4} - 2 x < \frac{π}{4}$ , which lies in $(- \frac{π}{2}, \frac{π}{2})$ .
So $tan^{- 1} (tan (\frac{π}{4} - 2 x)) = \frac{π}{4} - 2 x$ .

Also asked in: 2026 65/1/2, 2026 65/1/3

Q25 (OR) (OR)2 marksVery Short AnswerInverse Trigonometric Functions CBSE 2026 · 65/1/1

Evaluate : $tan (sin^{- 1} 1 - cos^{- 1} (- \frac{1}{2}))$

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Answer:

- \frac{1}{3}

$sin^{- 1} 1 = \frac{π}{2}$ and $cos^{- 1} (- \frac{1}{2}) = π - \frac{π}{3} = \frac{2 π}{3}$ .
$tan (\frac{π}{2} - \frac{2 π}{3}) = tan (- \frac{π}{6}) = - \frac{1}{3}$ .

Also asked in: 2026 65/1/2, 2026 65/1/3

Q232 marksVery Short AnswerInverse Trigonometric Functions CBSE 2026 · 65/3/1

Find the value of $sin [cot^{- 1} 2 (cos (tan^{- 1} 1))]$ .

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Answer:

\frac{1}{2}

$tan^{- 1} 1 = \frac{π}{4}$ , so $cos (tan^{- 1} 1) = \frac{1}{2}$ .
$2 \cdot \frac{1}{2} = 1$ , and $cot^{- 1} 1 = \frac{π}{4}$ .
$sin \frac{π}{4} = \frac{1}{2}$ .

Also asked in: 2026 65/3/2, 2026 65/3/3

Q212 marksVery Short AnswerInverse Trigonometric Functions CBSE 2026 · 65/4/1

Evaluate :
$tan^{- 1} (- \frac{1}{3}) + cot^{- 1} (\frac{1}{3}) + tan^{- 1} (sin (- \frac{π}{2})) + tan^{- 1} (tan \frac{2 π}{3})$

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Answer:

- \frac{5 π}{12}

$tan^{- 1} (- \frac{1}{3}) = - \frac{π}{6}$ .
$cot^{- 1} (\frac{1}{3}) = \frac{π}{3}$ .
$tan^{- 1} (sin (- \frac{π}{2})) = tan^{- 1} (- 1) = - \frac{π}{4}$ .
$tan \frac{2 π}{3} = - 3$ , so $tan^{- 1} (- 3) = - \frac{π}{3}$ (principal value).
Sum $= - \frac{π}{6} + \frac{π}{3} - \frac{π}{4} - \frac{π}{3} = - \frac{5 π}{12}$ .

Also asked in: 2026 65/4/2, 2026 65/4/3

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