## Trigonometry (11th Edition) Clone

$$\frac{1-\cos x}{1+\cos x}=\csc^2x-2\csc x\cot x+\cot^2x$$ The expression is proved to be an identity.
$$\frac{1-\cos x}{1+\cos x}=\csc^2x-2\csc x\cot x+\cot^2x$$ We look from the right side. $$A=\csc^2x-2\csc x\cot x+\cot^2x$$ We notice that the formula can be rewritten using the expansion $$(a-b)^2=a^2-2ab+b^2$$ Therefore, $$A=(\csc x-\cot x)^2$$ Now we take $$\csc x=\frac{1}{\sin x}\hspace{2cm}\cot x=\frac{\cos x}{\sin x}$$ That means $$\csc x-\cot x=\frac{1}{\sin x}-\frac{\cos x}{\sin x}=\frac{1-\cos x}{\sin x}$$ So, $$A=(\csc x-\cot x)^2=\frac{(1-\cos x)^2}{\sin^2x}$$ However, from Pythagorean Identities, we have $\sin^2x=1-\cos^2x=(1-\cos x)(1+\cos x)$. Therefore, $$A=\frac{(1-\cos x)^2}{(1-\cos x)(1+\cos x)}$$ $$A=\frac{1-\cos x}{1+\cos x}$$ The right side is therefore equal to the left side. The expression is an identity.