Prove this trig identity $$\sin^2(x) - \cos^2(x) - \tan^2(x) = \frac{2\sin^2(x) - 2\sin^4(x) - 1 }{ 1-\sin^2(x)}.$$
I tried this but I can't figure out how they got $-2\sin^4(x)$.

To get you started:
$$\begin{align}\sin^2 x - \cos^2 x - \tan^2 x & = \sin^2 x- \cos^2 x - \frac{ \sin^2 x }{\cos^2 x} \\ & = \frac{\sin^2 x\cos^2 - \cos^4 x - \sin^2 x}{\cos^2 x} \\ & = \frac{\sin^2 x (1-\sin^2 x)+(1-\sin^2 x)^2 - \sin^2 x}{1-\sin^2 x} & \\ & = \ldots \end{align}$$
Just expand and simplify.
