Prove that:

LHS $=\tan \left(\frac{\pi}{4}+x\right)+\tan \left(\frac{\pi}{4}-x\right)$

$=\frac{\tan \frac{\pi}{4}+\tan x}{1-\tan \frac{\pi}{4} \tan x}+\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \tan x} \quad\left[\because \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}\right.$ and $\left.\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}\right]$

$=\frac{1+\tan x}{1-\tan x}+\frac{1-\tan x}{1+\tan x}$

$=\frac{(1+\tan x)^{2}+(1-\tan x)^{2}}{(1+\tan x)(1-\tan x)}$

$=\frac{2\left(1+\tan ^{2} x\right)}{\left(1-\tan ^{2} x\right)}=\frac{2\left(\sec ^{2} x\right)}{1-\frac{\sin ^{2} x}{\cos ^{2} x}}$

$=\frac{2\left(\sec ^{2} x\right)\left(\cos ^{2} x\right)}{\cos 2 x} \quad\left(\because \cos ^{2} x-\sin ^{2} x=\cos 2 x\right)$

$=\frac{2 \times 1}{\cos 2 x}$

$=2 \sec 2 x=\mathrm{RHS}$

Hence proved.