Mathematics · 2019
JEE Main · 12 April 2019, Shift 2 · Q71
The derivative of tan^-1((sin x- cos x)/(sin x+ cos x)), with respect to x/2, where (x ∈(0, (π)/2)) is:
The derivative of $\displaystyle \tan ^{-1}\left(\frac{\sin x-\cos x}{\sin x+\cos x}\right)$,
with respect to $\displaystyle \frac{x}{2}$, where $\displaystyle \left(x \in\left(0, \frac{\pi}{2}\right)\right)$ is:
Official answer
(3)
$\displaystyle 2$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.