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....
$${\sin^2(x)\over \tan^2(x)} = \frac{\sin^2 x}{\frac{\sin^2 x}{\cos^2 x}}= \cos^2 x = 1-\sin^2 x$$
Notice that $\sec^2(x)=\frac{1}{\cos^2(x)}$ Putting it in our equation on R.H.S. it becomes:
 $${1-\cos^2(x)\over \sec^2(x)-1}={1-\cos^2(x)\over\frac{1}{\cos^2(x)}-1}$$
 $$=\frac{(1-\cos^2)(\cos^2(x))}{1-\cos^2(x)}$$
$$=\cos^2(x)=1-\sin^2(x)$$