Question

Prove that:
$\tan \left(\frac{\pi }{4}+\theta \right)-\tan \left(\frac{\pi }{4}-\theta \right)=2\tan 2\theta$.

Answer 1

STEP 1 : Solving the Left Hand Side (LHS) of the equation

Taking the LHS and solving we get,

$\tan \left(\frac{\pi }{4}+\theta \right)-\tan \left(\frac{\pi }{4}-\theta \right)$

$=\left[\frac{\tan \left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)+\tan \theta }{1-\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\tan \theta }\right]-\left[\frac{\tan \left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)-\tan \theta }{1+\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\tan \theta }\right]$

$=\left[\frac{1+\tan \theta }{1-\tan \theta }\right]-\left[\frac{1-\tan \theta }{1+\tan \theta }\right]$

$=\left[\frac{{\left(1+\tan \theta \right)}^2-{\left(1-\tan \theta \right)}^2}{\left(1-\tan \theta \right)\left(1+\tan \theta \right)}\right]$

$=\left[\frac{1^2+{\tan }^2\theta +2\times \tan \theta -\left(1^2+{\tan }^2\theta -2\times \tan \theta \right)}{1-{\tan }^2\theta }\right]$

$=\left[\frac{1+{\tan }^2\theta +2\tan \theta -1-{\tan }^2\theta +2\tan \theta }{1-{\tan }^2\theta }\right]$

$=\left[\frac{4\times \tan \theta }{1-{\tan }^2\theta }\right]=2\left[\frac{2\tan \theta }{1-{\tan }^2\theta }\right]$

$=2\tan 2\theta =RHS$

i.e. $\tan \left(\frac{\pi }{4}+\theta \right)-\tan \left(\frac{\pi }{4}-\theta \right)=2\tan 2\theta$

Hence proved.