Question

csc x - sin x
trigonometric identities

Answer 1

\[\csc x - \sin x=\frac{ 1 }{\sin x }-sinx =\frac{ 1-(\sin x)^2}{ \sin x } =\frac{ (cosx)^ 2}{ \sin x }\]
\[\frac{ (cosx)^2 }{ sinx }= \frac{cosx }{ sinx } \times \cos x =\cot x \times \cos x\]