#### Answer

$$\frac{\sin^4\alpha-\cos^4\alpha}{\sin^2\alpha-\cos^2\alpha}=1$$
The expression is an identity.

#### Work Step by Step

$$\frac{\sin^4\alpha-\cos^4\alpha}{\sin^2\alpha-\cos^2\alpha}=1$$
The left side is more complicated. We would simplify it.
$$A=\frac{\sin^4\alpha-\cos^4\alpha}{\sin^2\alpha-\cos^2\alpha}$$
We have $a^4-b^4=(a^2-b^2)(a^2+b^2)$. So,
$$A=\frac{(\sin^2\alpha-\cos^2\alpha)(\sin^2\alpha+\cos^2\alpha)}{\sin^2\alpha-\cos^2\alpha}$$
$$A=\sin^2\alpha+\cos^2\alpha$$
From Pythagorean Identity:
$$A=\sin^2\alpha+\cos^2\alpha=1$$
They are thus equal. The expression is an identity.