Question

Prove the following:

${\displaystyle \tan (\frac{\pi }{4}+\theta )=\frac{1+\tan \theta }{1-\tan \theta }}$

Answer 1

${\displaystyle L.H.S.=\tan (\frac{\pi }{4}+\theta )}$

According to ${\displaystyle \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}}$, 

We get

${\displaystyle L.H.S.=\frac{\tan \frac{\pi }{4}+\tan \theta }{1-\tan \frac{\pi }{4}\tan \theta }}$

${\displaystyle L.H.S=\frac{1+\tan \theta }{1-\left(1\right)\tan \theta }}$    ...[because ${\displaystyle \tan \frac{\pi }{4}=1]}$

=${\displaystyle \frac{1+\tan \theta }{1-\tan \theta }}$

Here ${\displaystyle L.H.S=R.H.S}$,  Hence proved