The derivative of $\tan ^{-1}\left(\frac{\sin x-\cos x}{\sin x+\cos x}\right)$, with respect to $\frac{x}{2}$, where $\left(x \in\left(0, \frac{\pi}{2}\right)\right)$ is:

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Differentiation

The derivative of

tan^{- 1} (\frac{s i n x - c o s x}{s i n x + c o s x})

, with respect to

\frac{x}{2}

, where

(x \in (0, \frac{π}{2}))

is:

Answer: (A)

Inside the inverse tangent, dividing top and bottom by

cos x

gives

\frac{t a n x - 1}{t a n x + 1} = tan (x - \frac{π}{4})

, so on

(0, \frac{π}{2})

the expression is simply

x - \frac{π}{4}

. Differentiating with respect to

x

gives

1

, but the derivative is wanted with respect to

\frac{x}{2}

, which is twice as large, namely

2

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option