## Trigonometry (11th Edition) Clone

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