## Trigonometry (11th Edition) Clone

$$\tan(\frac{\pi}{4}+x)=\frac{1+\tan x}{1-\tan x}$$
$$\tan(\frac{\pi}{4}+x)$$ Apply the identity of tangent of a sum here, we have $$=\frac{\tan\frac{\pi}{4}+\tan x}{1-\tan\frac{\pi}{4}\tan x}$$ $$=\frac{1+\tan x}{1-1\times \tan x}$$ $$=\frac{1+\tan x}{1-\tan x}$$