Solution:

$\tan \left(\frac{\pi}{4}+A\right)$

$=\frac{\tan \frac{\pi}{4}+\tan A}{1-\tan \frac{\pi}{4} \tan A}$

$=\frac{1+\tan \frac{\pi}{4} \tan A}{1-\tan \frac{\pi}{4}}$

$=\frac{1+\tan A}{1-\tan A}$

$=\frac{1+\frac{\sin A}{\cos A}}{1-\frac{\sin A}{\cos A}}$

$=\frac{\cos A+\sin A}{\cos A-\sin A}$