How do I prove $\csc^4 x-\cot^4x=(1+\cos^2x)/\sin^2x$

How do I prove $\csc^4 x-\cot^4x=\dfrac{(1+\cos^2x)}{\sin^2x}$

Do you start from RHS or LHS? I get stuck after first few steps-

• The right-hand side is currently not right. Our function on the left is $\frac{1-\cos^4 x}{\sin^4 x}$. But $1-\cos^4 x=(1-\cos^2 x)(1+\cos^2 x)=\sin^2 x(1+\cos^2 x)$. – André Nicolas May 11 '14 at 22:18
• ok I edited the identity, sorry for confusion... – Thetasquared May 11 '14 at 22:23
• hint: ${cos^2X}/{sin^2X} = cotX$ – Jason Chen May 11 '14 at 22:31
• are you sure it's $1+{cos^2X}/{sin^2X}$ and not ${(1+cos^2X)}/{sin^2X}$? – Jason Chen May 11 '14 at 22:33

The LHS: $$\frac{1}{\sin^4 x}-\frac{\cos^4 x}{\sin^4 x}=\frac{1-\cos^4 x}{\sin^4 x}=\frac{(1-\cos^2 x)(1+\cos^2 x)}{\sin^4 x}=\frac{1+\cos^2 x}{\sin^2 x}$$