Answer
$\tan(\frac{x}{2}+\frac{\pi}{4})=\sec x+\tan x$
Work Step by Step
Start with the left side:
$\tan(\frac{x}{2}+\frac{\pi}{4})$
Rewrite tangent as sine divided by cosine:
$=\frac{\sin (\frac{x}{2}+\frac{\pi}{4})}{\cos(\frac{x}{2}+\frac{\pi}{4})}$
Use addition formulas for sine and cosine:
$=\frac{\sin\frac{x}{2} \cos\frac{\pi}{4}+\cos\frac{x}{2} \sin\frac{\pi}{4}}{\cos\frac{x}{2} \cos\frac{\pi}{4}-\sin\frac{x}{2} \sin\frac{\pi}{4}}$
Evaluate $\sin\frac{\pi}{4}$ and $\cos\frac{\pi}{4}$:
$=\frac{\sin\frac{x}{2}*\frac{\sqrt{2}}{2}+\cos\frac{x}{2}*\frac{\sqrt{2}}{2}}{\cos\frac{x}{2} *\frac{\sqrt{2}}{2}-\sin\frac{x}{2} *\frac{\sqrt{2}}{2}}$
Divide top and bottom by $\frac{\sqrt{2}}{2}$:
$=\frac{\sin\frac{x}{2}+\cos\frac{x}{2}}{\cos\frac{x}{2}-\sin\frac{x}{2}}$
Multiply top and bottom by $\cos\frac{x}{2}+\sin\frac{x}{2}$:
$=\frac{\sin\frac{x}{2}+\cos\frac{x}{2}}{\cos\frac{x}{2}-\sin\frac{x}{2}}*\frac{\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{x}{2}+\sin\frac{x}{2}}$
$=\frac{\sin^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}+\cos^2\frac{x}{2}}{\cos^2\frac{x}{2}-\sin\frac{x}{2}}$
$=\frac{(\sin^2\frac{x}{2}+\cos^2\frac{x}{2})+2\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^2\frac{x}{2}-\sin\frac{x}{2}}$
Use double-angle formulas for sine and cosine to simplify the expression:
$=\frac{1+\sin(2*\frac{x}{2})}{\cos (2*\frac{x}{2})}$
$=\frac{1+\sin x}{\cos x}$
$=\frac{1}{\cos x}+\frac{\sin x}{\cos x}$
$=\sec x+\tan x$
