Prove this trig identity

Question

$$\sin^2(x) - \cos^2(x) - \tan^2(x) = \frac{2\sin^2(x) - 2\sin^4(x) - 1 }{ 1-\sin^2(x)}.$$

I tried this but I can't figure out how they got $-2\sin^4(x)$.

Answer 1

\begin{align*} \sin^2\theta -\cos^2\theta-\tan^2\theta & = \sin^2\theta -\cos^2\theta-\frac{\sin^2\theta}{\cos^2\theta}\\ & = \frac{\sin^2\theta \cos^2\theta -\cos^4\theta-\sin^2\theta}{\cos^2\theta}\\ & = \frac{\sin^2\theta (1-\sin^2\theta) -(1-\sin^2\theta)^2-\sin^2\theta}{\cos^2\theta}\\ \end{align*} Now simplify it.

Answer 2

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.

Answer 3

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