Mathematics · 2019
JEE Main · 8 April 2019, Shift 1 · Q71
If 2 y=( cot^-1((√ 3 cos x+ sin x)/(cos x-√ 3 sin x)))^2, x ∈(0, (π)/2) then (d y)/(d x) is equal to:
If
$$2 y=\left(\cot ^{-1}\left(\frac{\sqrt{3} \cos x+\sin x}{\cos x-\sqrt{3} \sin x}\right)\right)^2, x \in\left(0, \frac{\pi}{2}\right)
$$
then $\displaystyle \frac{\mathrm{d} y}{\mathrm{~d} x}$ is equal to :
Official answer
(3)
$\displaystyle x-\frac{\pi}{6}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.