Trigonometry (11th Edition) Clone

$$\frac{1+\cos x}{1-\cos x}-\frac{1-\cos x}{1+\cos x}=4\cot x\csc x$$ The expression is an identity. We simplify the left side.
$$\frac{1+\cos x}{1-\cos x}-\frac{1-\cos x}{1+\cos x}=4\cot x\csc x$$ We simplify the left side first, since it is obviously more complex-looking. $$A=\frac{1+\cos x}{1-\cos x}-\frac{1-\cos x}{1+\cos x}$$ $$A=\frac{(1+\cos x)^2-(1-\cos x)^2}{(1-\cos x)(1+\cos x)}$$ (we know that $a^2-b^2=(a-b)(a+b)$) $$A=\frac{[(1+\cos x)-(1-\cos x)][(1+\cos x)+(1-\cos x)]}{1-\cos^2 x}$$ $$A=\frac{[1+\cos x-1+\cos x][1+\cos x+1-\cos x]}{1-\cos^2 x}$$ $$A=\frac{2\cos x\times2}{1-\cos^2x}$$ $$A=\frac{4\cos x}{\sin^2 x}$$ (from Pythagorean Identity: $1-\cos^2x=\sin^2x$) $$A=4\frac{\cos x}{\sin x}\frac{1}{\sin x}$$ We would apply the identities: $\frac{\cos x}{\sin x}=\cot x$ and $\frac{1}{\sin x}=\csc x$. That means, $$A=4\cot x\csc x$$ The expression is hence proved to be an identity, since both sides are equal.