Answer
$$\frac{1}{1+\cos x}-\frac{1}{1-\cos x}=-2\cos x\csc^2 x$$
Work Step by Step
$$A=\frac{1}{1+\cos x}-\frac{1}{1-\cos x}$$
$$A=\frac{1-\cos x}{(1+\cos x)(1-\cos x)}-\frac{1+\cos x}{(1+\cos x)(1-\cos x)}$$
$$A=\frac{1-\cos x-(1+\cos x)}{(1+\cos x)(1-\cos x)}$$
$$A=\frac{1-\cos x-1-\cos x}{(1+\cos x)(1-\cos x)}$$
$$A=\frac{-2\cos x}{(1+\cos x)(1-\cos x)}$$
$$A=\frac{-2\cos x}{1-\cos^2 x}$$ (for $a^2-b^2=(a-b)(a+b)$)
- Pythagorean Identity:
$$\sin^2 x=1-\cos^2 x$$
Therefore, $$A=\frac{-2\cos x}{\sin^2 x}$$
$$A=-2\cos x(\frac{1}{\sin x})^2$$
- Reciprocal Identity:
$$\csc x=\frac{1}{\sin x}$$
So, $$A=-2\cos x\csc^2 x$$