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 $-2\sin^4(x)$.
\begin{align*}
\sin^2\theta -\cos^2\theta-\tan^2\theta &amp; = \sin^2\theta -\cos^2\theta-\frac{\sin^2\theta}{\cos^2\theta}\\
&amp; = \frac{\sin^2\theta \cos^2\theta -\cos^4\theta-\sin^2\theta}{\cos^2\theta}\\
&amp; = \frac{\sin^2\theta (1-\sin^2\theta) -(1-\sin^2\theta)^2-\sin^2\theta}{\cos^2\theta}\\
\end{align*}
Now simplify it.
you can factor sin^2 theta from 2sin^2-2sin^4 

then put 1-sin^2 theta as cos ^2 theta 

write tan^2 theta as sin^2/cos^2

then put cos ^2 as 1-sin^2 theta