Answer
$\dfrac{1+\tan x}{1-\tan x}=\dfrac{\cos x+\sin x}{\cos x-\sin x}$
Work Step by Step
$\dfrac{1+\tan x}{1-\tan x}=\dfrac{\cos x+\sin x}{\cos x-\sin x}$
Replace $\tan x$ by $\dfrac{\sin x}{\cos x}$:
$\dfrac{1+\dfrac{\sin x}{\cos x}}{1-\dfrac{\sin x}{\cos x}}=\dfrac{\cos x+\sin x}{\cos x-\sin x}$
Evaluate the operations indicated in the numerator and the denominator of the fraction on the left:
$\dfrac{\dfrac{\cos x+\sin x}{\cos x}}{\dfrac{\cos x-\sin x}{\cos x}}=\dfrac{\cos x+\sin x}{\cos x-\sin x}$
Evaluate the division and the identity will be proved:
$\dfrac{\cos x(\cos x+\sin x)}{\cos x(\cos x-\sin x)}=\dfrac{\cos x+\sin x}{\cos x-\sin x}$
$\dfrac{\cos x+\sin x}{\cos x-\sin x}=\dfrac{\cos x+\sin x}{\cos x-\sin x}$
