Question

Prove that
$\tan (\pi /4+\theta )-\tan (\pi /4-\theta )=2\tan 2\theta$

Answer 1

$\tan (\frac{\pi }{4}+\theta )=\frac{\tan \left(\pi /4\right)+\tan \theta }{1-\tan \left(\pi /4\right)\tan \theta }=\frac{1+\tan \theta }{1-\tan \theta }$
Similarly, $\tan (\frac{\pi }{4}-\theta )=\frac{1+\tan \theta }{1-\tan \theta }$
LHS $=\frac{1+\tan \theta }{1-\tan \theta }-\frac{1-\tan \theta }{1+\tan \theta }=\frac{{(1+\tan \theta )}^2-{(1-\tan \theta )}^2}{1-{\tan }^2\theta }=\frac{4\tan \theta }{1-{\tan }^2\theta }=2.\frac{2\tan \theta }{1-{\tan }^2\theta }=2\tan 2\theta$