## Trigonometry (11th Edition) Clone

$$\frac{\sin\theta}{1-\cos\theta}-\frac{\sin\theta\cos\theta}{1+\cos\theta}=\csc\theta(1+\cos^2\theta)$$ The expression is proved to be an identity by simplifying the left side.
$$\frac{\sin\theta}{1-\cos\theta}-\frac{\sin\theta\cos\theta}{1+\cos\theta}=\csc\theta(1+\cos^2\theta)$$ 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.