Question

Differentiate,${\tan }^{-1}\frac{(\cos x+\sin x}{(\cos x-\sin x)})$ with respect to $x$

• A. $1$
• B. $0$
• C. $\frac{1}{2}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is A

$1$

Differentiate with respect to $\mathit{X}$

$$\begin{gathered} Lety=\frac{\tan -1(\cos x+\sin x)}{(\cos x–\sin x)} \\ ={\tan }^{-1}\frac{(1+\tan x)}{(1–\tan x)} \\ ={\tan }^{-1}\frac{(\tan \frac{\pi }{4}+\tan x)}{(\tan \frac{\pi }{4}–\tan \frac{\pi }{4}\tan x)} \\ ={\tan }^{-1}\tan (\frac{\pi }{4}+x) \\ =(\frac{\pi }{4}+x) \\ \frac{dy}{dx}=0+1 \end{gathered}$$

$\mathbf{=}\mathbf{1}$

Hence, Differentiation of ${\tan }^{-1}\frac{(\cos x+\sin x}{(\cos x-\sin x)})$ w.r.t $\mathit{X}$ is $\mathbf{1}$.