Answer
$csc^{4}x-\cot ^{4}x=\dfrac {1+\cos ^{2}x}{1-\cos ^{2}x} $
Work Step by Step
$csc^{4}x-\cot ^{4}x=\dfrac {1}{\sin ^{4}x}-\dfrac {\cos ^{4}x}{\sin ^{4}x}=\dfrac {1-\left( \cos ^{2}x\right) ^{2}}{\sin ^{4}x}=\dfrac {\left( 1-\cos ^{2}x\right) \left( 1+\cos ^{2}x\right) }{\sin ^{4}x}=\dfrac {\sin ^{2}x\left( 1+\cos ^{2}x\right) }{\sin ^{4}x}=\dfrac {1+\cos ^{2}x}{\sin ^{2}x}=\dfrac {1+\cos ^{2}x}{1-\cos ^{2}x} $