#### Answer

$$\frac{\sin^2\theta-\cos^2\theta}{\sin^4\theta-\cos^4\theta}=1$$
The equation is an identity as 2 sides are equal to each other.

#### Work Step by Step

$$\frac{\sin^2\theta-\cos^2\theta}{\sin^4\theta-\cos^4\theta}=1$$
Let's consider the left side.
$$A=\frac{\sin^2\theta-\cos^2\theta}{\sin^4\theta-\cos^4\theta}$$
We see that both numerator and denominator can be expanded using the expansion $X^2-Y^2=(X-Y)(X+Y)$ and $X^4-Y^4=(X^2-Y^2)(X^2+Y^2)$
However, it might be wiser to only expand the denominator now, as you will see later.
$$A=\frac{\sin^2\theta-\cos^2\theta}{(\sin^2\theta-\cos^2\theta)(\sin^2\theta+\cos^2\theta)}$$
$$A=\frac{1}{\sin^2\theta+\cos^2\theta}$$
So by only expanding the denominator, the whole numerator has been simplified.
Here, we apply Pythagorean Identity: $\sin^2\theta+\cos^2\theta=1$.
$$A=\frac{1}{1}=1$$
Thus, both sides are equal to each other, meaning the equation is an identity.