CBSE Maths › Class 12 PYQs › Inverse Trigonometric Functions

Inverse Trigonometric Functions: 2 marks Questions (CBSE Class 12)

11 different 2 marks questions on Inverse Trigonometric Functions from CBSE Class 12 Maths board exams 2026, newest first.

1 mark (11)2 marks (11)5 marks (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

Q252 marksVery Short AnswerInverse Trigonometric Functions CBSE 2026 · 65/1/2

Simplify : $cot^{- 1} \frac{1 + c o s 2 x}{1 - c o s 2 x}, x \in (0, \frac{π}{2})$ .

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Answer:

x

$\frac{1 + c o s 2 x}{1 - c o s 2 x} = \frac{2 c o s ^{2} x}{2 s i n ^{2} x} = cot^{2} x$ .
For $x \in (0, \frac{π}{2})$ , $cot x > 0$ , so the square root is $cot x$ .
$cot^{- 1} (cot x) = x$ since $x \in (0, π)$ .

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

Evaluate : $sin (tan^{- 1} (- 3) - sec^{- 1} 2)$

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Answer:

- \frac{3}{2}

$tan^{- 1} (- 3) = - \frac{π}{3}$ and $sec^{- 1} 2 = \frac{π}{3}$ (principal values).
$sin (- \frac{π}{3} - \frac{π}{3}) = sin (- \frac{2 π}{3}) = - \frac{3}{2}$ .

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

Simplify : $sin^{- 1} \frac{1 + c o s 2 x}{2}, 0 < x < \frac{π}{2}$ .

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Answer:

\frac{π}{2} - x

$\frac{1 + c o s 2 x}{2} = cos^{2} x$ , and $cos x > 0$ for $0 < x < \frac{π}{2}$ , so the square root is $cos x$ .
$sin^{- 1} (cos x) = sin^{- 1} (sin (\frac{π}{2} - x))$ .
Since $0 < \frac{π}{2} - x < \frac{π}{2}$ , this equals $\frac{π}{2} - x$ .

Q23 (OR) (OR)2 marksVery Short AnswerInverse Trigonometric Functions CBSE 2026 · 65/1/3

Evaluate : $cos [sin^{- 1} (- 1) - tan^{- 1} (- 3)]$ .

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Answer:

\frac{3}{2}

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

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

Q212 marksVery Short AnswerInverse Trigonometric Functions CBSE 2026 · 65/5/1

Evaluate $sin [tan^{- 1} tan (\frac{3 π}{4})]$ .

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Answer:

- \frac{1}{2}

$tan \frac{3 π}{4} = - 1$ .
$tan^{- 1} (- 1) = - \frac{π}{4}$ (principal value in $(- \frac{π}{2}, \frac{π}{2})$ ).
$sin (- \frac{π}{4}) = - \frac{1}{2}$ .

Q212 marksVery Short AnswerInverse Trigonometric Functions CBSE 2026 · 65/5/2

Evaluate $sin [cos^{- 1} cos (\frac{7 π}{6})]$ .

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Answer:

\frac{1}{2}

$cos \frac{7 π}{6} = - \frac{3}{2}$ .
$cos^{- 1} (- \frac{3}{2}) = \frac{5 π}{6}$ (principal value in $[0, π]$ ).
$sin \frac{5 π}{6} = \frac{1}{2}$ .

Q212 marksVery Short AnswerInverse Trigonometric Functions CBSE 2026 · 65/5/3

Evaluate $tan [cos^{- 1} (tan \frac{3 π}{4})]$ .

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Answer: 0

$tan \frac{3 π}{4} = - 1$ .
$cos^{- 1} (- 1) = π$ (principal value in $[0, π]$ ).
$tan π = 0$ .

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