Trigonometry identity $\csc x\cot x=\frac{\cos ^3x}{\sin^2 x}+\cos x$ How to prove that $\csc x\cot x=\frac{\cos ^3x}{\sin^2 x}+\cos x$?
I tried manupulating the left hand side but ended up in $\frac{\cos x}{\sin^2 x}$.
Can someone show me? Thanks in advance.

$RHS = \dfrac{\cos^3 x}{\sin^2 x}+ \cos x=\dfrac{\cos^3 x+\cos x \sin^2x}{\sin^2 x}=\dfrac{\cos x (\sin^2x+\cos^2 x)}{\sin^2 x}=\dfrac{\cos x}{\sin^2x}$ BECAUSE $\sin^2 x + \cos^2 x =1$...
Now, $\dfrac{\cos x}{\sin^2x}=\dfrac{\cos x}{\sin x} \dfrac{1}{\sin x}=\cot x \csc x.$
