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JEE Advanced 2026 · Paper 1, Mathematics question 4

Considering only the principal values of the inverse trigonometric functions, the value of
cot⁡−1(cot⁡(−11))+10sin⁡(2cos⁡−1(12))+10sin⁡(2tan⁡−1(2))\cot^{-1}(\cot(-11))+10\sin\left(2\cos^{-1}\left(\frac1{\sqrt2}\right)\right)+10\sin\left(2\tan^{-1}(2)\right)
is

Choose one option
Answer and worked solution

Official answer: C

First term. cot⁡−1\cot^{-1} takes values in (0,π)(0,\pi). Since cot⁡\cot has period π\pi, cot⁡(−11)=cot⁡(4π−11)\cot(-11)=\cot(4\pi-11), and 4π−11≈1.566∈(0,π)4\pi-11\approx1.566\in(0,\pi). So the term is 4π−114\pi-11.

Second term. 2cos⁡−112=π22\cos^{-1}\frac1{\sqrt2}=\frac\pi2, so 10sin⁡π2=1010\sin\frac\pi2=10.

Third term. sin⁡(2tan⁡−12)=2⋅21+4=45\sin(2\tan^{-1}2)=\dfrac{2\cdot2}{1+4}=\dfrac45, so the term is 88.

Total. 4π−11+10+8=4π+74\pi-11+10+8=4\pi+7.

Topics:Inverse trigonometric functions
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