Question

Solve $ta{n}^{-1}\left\{\frac{cosx+sinx}{cosx-sinx}\right\}$

Answer 1

${\tan }^{-1}\left\{\frac{\cos x+\sin x}{\cos x-\sin x}\right\}$

$={\tan }^{-1}\left\{\frac{1+\frac{\sin x}{\cos x}}{1-\frac{\sin x}{\cos x}}\right\}$ by dividing the numerator and denominator by $\cos x$

$={\tan }^{-1}\left\{\frac{\tan \frac{\pi }{4}+\tan x}{1-\tan \frac{\pi }{4}\tan x}\right\}$ where $1=\tan \frac{\pi }{4}$

$={\tan }^{-1}\tan (\frac{\pi }{4}+x)$ since $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

$=\frac{\pi }{4}+x$