Question

Differentiate

${\tan }^{-1}\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right),\frac{\pi }{4}<x<\frac{\pi }{4}$

Answer 1

$$\begin{gathered} \mathrm{Let},y={\tan }^{-1}\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right) \\ \Rightarrow y={\tan }^{-1}\left[\frac{{\displaystyle \frac{\cos x+\sin x}{\cos x}}}{{\displaystyle \frac{\cos x-\sin x}{\cos x}}}\right] \\ \Rightarrow y={\tan }^{-1}\left[\frac{{\displaystyle \frac{\cos x}{\cos x}}+{\displaystyle \frac{\sin x}{\cos x}}}{{\displaystyle \frac{\cos x}{\cos x}}-{\displaystyle \frac{\sin x}{\cos x}}}\right] \\ \Rightarrow y={\tan }^{-1}\left[\frac{1+\tan x}{1-\tan x}\right] \\ \Rightarrow y={\tan }^{-1}\left[\frac{\tan {\displaystyle \frac{\pi }{4}}+\tan x}{1-{\displaystyle \tan \frac{{\displaystyle \pi }}{4}}\tan x}\right] \\ \Rightarrow y={\tan }^{-1}\left[\tan \left(\frac{\mathrm{\pi }}{4}+x\right)\right] \\ \Rightarrow y=\frac{\mathrm{\pi }}{4}+x \end{gathered}$$

Differentiate it with respect to x,

$$\begin{gathered} \frac{dy}{dx}=0+1 \\ \therefore \frac{dy}{dx}=1 \end{gathered}$$