## Trigonometry (11th Edition) Clone

$tan~(\frac{x}{2}+\frac{\pi}{4}) = sec~x+tan~x$
We can verify that the equation is an identity: $tan~(\frac{x}{2}+\frac{\pi}{4})$ $= \frac{sin~(\frac{x}{2}+\frac{\pi}{4})}{cos~(\frac{x}{2}+\frac{\pi}{4})}$ $= \frac{sin~\frac{x}{2}~cos~\frac{\pi}{4}+cos~\frac{x}{2}~sin~\frac{\pi}{4}}{cos~\frac{x}{2}~cos~\frac{\pi}{4}-sin~\frac{x}{2}~sin~\frac{\pi}{4}}$ $= \frac{sin~\frac{x}{2}~(\frac{\sqrt{2}}{2})+cos~\frac{x}{2}~(\frac{\sqrt{2}}{2})}{cos~\frac{x}{2}~(\frac{\sqrt{2}}{2})-sin~\frac{x}{2}~(\frac{\sqrt{2}}{2})}$ $= \frac{sin~\frac{x}{2}+cos~\frac{x}{2}}{cos~\frac{x}{2}-sin~\frac{x}{2}}$ $= \frac{sin~\frac{x}{2}+cos~\frac{x}{2}}{cos~\frac{x}{2}-sin~\frac{x}{2}}~\times~ \frac{sin~\frac{x}{2}+cos~\frac{x}{2}}{sin~\frac{x}{2}+cos~\frac{x}{2}}$ $= \frac{sin^2~\frac{x}{2}+cos^2~\frac{x}{2}+2~sin~\frac{x}{2}~cos~\frac{x}{2}}{cos^2~\frac{x}{2}-sin^2~\frac{x}{2}}$ $= \frac{1+2~sin~\frac{x}{2}~cos~\frac{x}{2}}{cos^2~\frac{x}{2}-sin^2~\frac{x}{2}}$ $= \frac{1+sin~x}{cos~x}$ $= \frac{1}{cos~x}+\frac{sin~x}{cos~x}$ $= sec~x+tan~x$