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

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

Answer: 
For the equation: $$\frac{1-\cos^2x}{\sec^2x-1}$$

Multiply the numerator and denominator by $\cos^2x$ 

We now get: $$\frac{\cos^2x-\cos^4x}{1-\cos^2x}$$

Separate this into two fractions: $$\frac{\cos^2x}{\sin^2x} - \frac{\cos^4x}{\sin^2x}$$

This can then be converted to: $$\cot^2x - \cot^2x\cos^2x$$

We take $\cot^2x$ common, and get: $\cot^2x\sin^2x$

When this is multiplied, this gives us: $\cos^2x$, or rather, $1-\sin^2x$

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