Answer
$\tan\Big(x+\dfrac{\pi}{4}\Big)=\dfrac{1+\tan x}{1-\tan x}$
Work Step by Step
$\tan\Big(x+\dfrac{\pi}{4}\Big)=\dfrac{1+\tan x}{1-\tan x}$
Apply the addition formula for the tangent on the left side:
$\dfrac{\tan x+\tan\dfrac{\pi}{4}}{1-\tan x\tan\dfrac{\pi}{4}}=\dfrac{1+\tan x}{1-\tan x}$
Evaluate $\tan\dfrac{\pi}{4}$:
$\dfrac{\tan x+1}{1-(1)\tan x}=\dfrac{1+\tan x}{1-\tan x}$
Simplify and the identity will be proved:
$\dfrac{1+\tan x}{1-\tan x}=\dfrac{1+\tan x}{1-\tan x}$
