Question

Prove the following :
$\tan (\frac{\pi }{4}+\theta )=\frac{1+\tan \theta }{1-\tan \theta }$

Answer 1

we know $\tan (a+b)=\frac{\tan \left(a\right)+\tan \left(b\right)}{1-\tan \left(a\right).\tan \left(b\right)}$

given $\tan (\frac{\pi }{4}+\theta )=\frac{\tan \frac{\pi }{4}+\tan \theta }{1-\tan \frac{\pi }{4}\times \tan \theta }\Rightarrow =\frac{1+\tan \theta }{1-\tan \theta }[\because \tan \frac{\pi }{4}=1]$[using above formula]