#### Answer

$$\frac{\sec^4\theta-\tan^4\theta}{\sec^2\theta+\tan^2\theta}=\sec^2\theta-\tan^2\theta$$
We simplify the left side and find that the expression is an identity.

#### Work Step by Step

$$\frac{\sec^4\theta-\tan^4\theta}{\sec^2\theta+\tan^2\theta}=\sec^2\theta-\tan^2\theta$$
The left side is more complicated. We would simplify it.
$$A=\frac{\sec^4\theta-\tan^4\theta}{\sec^2\theta+\tan^2\theta}$$
We have $a^4-b^4=(a^2-b^2)(a^2+b^2)$. So,
$$A=\frac{(\sec^2\theta-\tan^2\theta)(\sec^2\theta+\tan^2\theta)}{\sec^2\theta+\tan^2\theta}$$
$$A=\sec^2\theta-\tan^2\theta$$
They are thus equal. The expression is an identity.