Prove that $\frac{1}{1-\sin x}-\frac{1}{1+\sin x}=2 \tan x \sec x$

Taking a common denominator in the left hand side we get
$$\frac{1}{1-\sin x}-\frac{1}{1+\sin x}=\frac{1+\sin x -(1-\sin x)}{1-\sin ^2 x}=\frac{2 \sin x}{\cos^2 x}$$
\[=2\frac{\sin x}{\cos x} \times \frac{1}{\cos x}=2 \tan x \times \sec x,\]
and we are done.