Question: ${\sin^2(x)\over \tan^2(x)}$

I did this and then got stuck. Could someone give me some hints please?

Answer: 
Good job at arriving at 
$${1-\cos^2(x)\over \sec^2(x)-1}= {\sin^2(x)\over \tan^2(x)}$$ 
We know that $\tan x = \dfrac {\sin x}{\cos x}.$ So....
$${\sin^2(x)\over \tan^2(x)} = \frac{\sin^2 x}{\frac{\sin^2 x}{\cos^2 x}}= \cos^2 x = 1-\sin^2 x$$
we have $$\frac{1}{\sec(x)^2-1}=\frac{\cos(x)^2}{1-\cos(x)^2}$$ and from both we get $$\cos(x)^2=1-\sin(x)^2$$