Answer
$\frac{1-sin^2x}{csc^2x-1}=sin^2x$
Work Step by Step
We know that:
$sin^2x+cos^2x=1$
$cos^2x=1-sin^2x$
$csc^2x=1+cot^2x$
$csc^2x-1=cot^2x$
$cot~x=\frac{cos~x}{sin~x}$
So:
$\frac{1-sin^2x}{csc^2x-1}=\frac{cos^2x}{cot^2x}=\frac{cos^2x}{(\frac{cos~x}{sin~x})^2}=cos^2x~\frac{sin^2x}{cos^2x}=sin^2x$
