Show that ${1-\cos^2(x)\over \sec^2(x)-1}=1-\sin^2(x)$ ${\sin^2(x)\over \tan^2(x)}$
I did this and then got stuck. Could someone give me some hints please?
 A: 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....
$${\sin^2(x)\over \tan^2(x)} = \frac{\sin^2 x}{\frac{\sin^2 x}{\cos^2 x}}= \cos^2 x = 1-\sin^2 x$$
A: 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.
A: we have $$\frac{1}{\sec(x)^2-1}=\frac{\cos(x)^2}{1-\cos(x)^2}$$ and from both we get $$\cos(x)^2=1-\sin(x)^2$$
A: Notice that $\sec^2(x)=\frac{1}{\cos^2(x)}$ Putting it in our equation on R.H.S. it becomes:
  $${1-\cos^2(x)\over \sec^2(x)-1}={1-\cos^2(x)\over\frac{1}{\cos^2(x)}-1}$$
   $$=\frac{(1-\cos^2)(\cos^2(x))}{1-\cos^2(x)}$$
$$=\cos^2(x)=1-\sin^2(x)$$
