Question

prove this identity:
(cosx + 1) / (sin^3x) = (cscx) / (1 - cosx)

Answer 1

\[\frac{\cos x + 1}{\sin^3 x} = \left(\frac{\cos x + 1}{\sin^3 x}\right)\left( \frac{\cos x - 1}{\cos x - 1} \right)\]\[= \frac{\cos^2 x - 1}{\left(\sin^3 x\right)\left(\cos x - 1\right)}\]\[= \frac{\left( 1 - \sin^2 x \right) - 1}{\left(\sin^3 x\right)\left(\cos x - 1\right)}\]\[= \frac{-\cancel{\sin^2 x}}{\left(\cancel{\sin^2 x}\right)\left(\sin x\right)\left(\cos x - 1\right)}\]\[= \frac{-1}{\left(\sin x\right)\left(\cos x - 1\right)}\]\[= \frac{\csc x}{1 - \cos x}\]