$\tan(\large\frac{\pi}{4}$$+\theta)+\tan(\large\frac{\pi}{4}-$$\theta)$
$\large\frac{\tan\Large\frac{\pi}{4}+\tan\theta}{1-\tan \Large\frac{\pi}{4}.\tan\theta}+\frac{\tan\Large\frac{\pi}{4}-\tan \theta}{1+\tan \Large\frac{\pi}{4}.\tan\theta}$
$\Rightarrow \large\frac{1+\tan \theta}{1-\tan \theta}+\large\frac{1-\tan \theta}{1+\tan \theta}$
$\Rightarrow \large\frac{(1+\tan \theta)^2+(1-\tan \theta)^2}{(1-\tan \theta)(1+\tan \theta)}$
$\Rightarrow \large\frac{(1+\tan^2\theta+2\tan \theta)+(1+\tan^2\theta-2\tan \theta)}{1-\tan^2\theta}$
$\Rightarrow \large\frac{2(1+\tan^2\theta)}{1-\tan^2\theta}$
$\Rightarrow \large\frac{2}{\cos 2\theta}$
$\Rightarrow \sec 2\theta$=RHS
Hence proved