Answer
$$\frac{\sin\theta}{1-\cos\theta}-\frac{\sin\theta\cos\theta}{1+\cos\theta}=\csc\theta(1+\cos^2\theta)$$
The expression is proved to be an identity by simplifying the left side.
Work Step by Step
$$\frac{\sin\theta}{1-\cos\theta}-\frac{\sin\theta\cos\theta}{1+\cos\theta}=\csc\theta(1+\cos^2\theta)$$
The left side is more complicated, which means we would try to simplify it first.
$$A=\frac{\sin\theta}{1-\cos\theta}-\frac{\sin\theta\cos\theta}{1+\cos\theta}$$
$$A=\frac{\sin\theta(1+\cos\theta)-\sin\theta\cos\theta(1-\cos\theta)}{(1+\cos\theta)(1-\cos\theta)}$$
$$A=\frac{\sin\theta+\sin\theta\cos\theta-\sin\theta\cos\theta+\sin\theta\cos^2\theta}{1-\cos^2\theta}$$
$$A=\frac{\sin\theta+\sin\theta\cos^2\theta}{1-\cos^2\theta}$$
$$A=\frac{\sin\theta(1+\cos^2\theta)}{1-\cos^2\theta}$$
- From Pythagorean Identity, we have $$1-\cos^2\theta=\sin^2\theta$$. So, $$A=\frac{\sin\theta(1+\cos^2\theta)}{\sin^2\theta}$$
$$A=\frac{1+\cos^2\theta}{\sin\theta}$$
$$A=\frac{1}{\sin\theta}(1+\cos^2\theta)$$
- From Reciprocal Identity: $$\frac{1}{\sin\theta}=\csc\theta$$. So,
$$A=\csc\theta(1+\cos^2\theta)$$
That makes the left and right sides equal with each other. The expression is thus an identity.
