JEE Advanced 2026 Paper 1, Mathematics Section 1 Q4: Inverse Trigonometric Functions

JEE Advanced2026Paper 1Mathematics Section 1

Considering only the principal values of the inverse trigonometric functions, the value of

$cot^{- 1} (cot (- 11)) + 10 sin (2 cos^{- 1} (\frac{1}{2})) + 10 sin (2 tan^{- 1} (2))$

A
$3 π + 7$
B
$7$
C
$4 π + 7$
D
$3 π - 5$

Solution

First term: The principal range of $cot^{- 1}$ is $(0, π)$ . We need $4 π - 11 \in (0, π)$ , which is true since $11 \approx 3.5 π$ lies between $3 π$ and $4 π$ . Using $cot (- 11) = cot (4 π - 11)$ , we get $cot^{- 1} (cot (- 11)) = 4 π - 11$ .

Second term: $cos^{- 1} (1/ 2) = π /4$ , so $10 sin (2 \cdot π /4) = 10 sin (π /2) = 10$ .

Third term: Let $θ = tan^{- 1} (2)$ with $θ \in (0, π /2)$ and $tan θ = 2$ . Then

$sin 2 θ = \frac{2 tan θ}{1 + tan ^{2} θ} = \frac{4}{5}$

So the third term equals $10 \cdot \frac{4}{5} = 8$ .

Total: $(4 π - 11) + 10 + 8 = 4 π + 7$ .

$∴$ The correct option is (C).

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JEE Advanced 2026 Paper 1, Mathematics Section 1 Q4: Inverse Trigonometric Functions

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