The derivative of tan⁻¹((cos x - sin x)/(cos x + sin x)) with respect to x, where x ∈ (0, π/2) is:

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

(a)$\dfrac{\pi}{4} + 1$
(b)$1$
(c)$\dfrac{\pi}{4}$
(d)$-1$

Answer

Correct option: (d)

Answer: (d) $-1$ $\dfrac{\cos x - \sin x}{\cos x + \sin x} = \dfrac{1 - \tan x}{1 + \tan x} = \tan\left(\dfrac{\pi}{4} - x\right)$, so the function is $\dfrac{\pi}{4} - x$ for $x \in \left(0, \dfrac{\pi}{2}\right)$ and its derivative is $-1$.