Question

Prove identity:
cosx+1/sin^3x = cscx/1-cosx

Answer 1

(1+cosx/sin^2x) .sinx=1+cosx/1-cos^2x .coscx=coscx/1-cosx

Answer 2

\[\large{\frac{cosx+1}{\sin^3x}}\]\[\large{=\frac{cosx+1}{\sin^2x \times sinx}=\frac{cosx+1}{(1-\cos^2x)sinx}}\]\[\large{=\frac{\cancel{cosx+1}}{(1-cosx) \cancel{(1+cosx)} sinx}}\]\[=\frac{1}{(1-cosx)sinx}\]\[\large{=\frac{1}{1-cosx} \times \frac{1}{sinx}}\]and 1/sinx=cosecx
so
\[\large{=\frac{1}{1-cosx}\times cosecx}\]\[\large{=\frac{cosecx}{1-cosx}}\]