Answer
$\frac{sin~x}{1+cos~x}+\frac{sin~x}{1-cos~x}=2~csc~x$
Work Step by Step
$\frac{sin~x}{1+cos~x}+\frac{sin~x}{1-cos~x}=\frac{sin~x}{1+cos~x}\frac{1-cos~x}{1-cos~x}+\frac{sin~x}{1-cos~x}\frac{1+cos~x}{1+cos~x}=\frac{sin~x-sin~x~cos~x+sin~x+sin~x~cos~x}{1-cos^2x}=\frac{2~sin~x}{sin^2x}=\frac{2}{sin~x}=2~csc~x$
