Question

sinx/1-cosx = 1+cosx/sinx

Answer 1

\[\large \frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}\]

Answer 2

Multiply both sides by sinx, then multiply both sides by 1-cosx (ie cross multiply the fractions)
\[\large \sin^2 x = (1+\cos x) (1-\cos x)\]
\[\large \sin^2 x = 1- \cos^2 x\]

Remember that \[\large 1 = \sin^2 x + \cos^2 x \]

Answer 3

\[ \frac{1+\cos x}{\sin x}\] or let's just work on the right hand side and leave the left side alone. Multiply the fraction by (1-cosx)/(1-cosx)

\[ \frac{(1+\cos x)(1-\cos x)}{\sin x(1-cosx)}\]
\[\frac{1-\cos^2 x}{\sin x(1-\cos x})\]

\[\frac{ \sin^2 x }{ \sin x (1-\cos x) }\] which if you simplify a little, you'll get the left hand side.