#### Answer

$$\frac{\cos^4x-\sin^4x}{\cos^2x}=1-\tan^2x$$
The detailed step of verification is in the Work step by step.

#### Work Step by Step

$$\frac{\cos^4x-\sin^4x}{\cos^2x}=1-\tan^2x$$
We start from the left side one more time.
$$X=\frac{\cos^4x-\sin^4x}{\cos^2x}$$
$$X=\frac{(\cos^2x)^2-(\sin^2x)^2}{\cos^2x}$$
$$X=\frac{(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)}{\cos^2x}$$ ($A^2-B^2=(A-B)(A+B)$)
- $\cos^2x+\sin^2x$ is equal to $1$, so we write it into $1$.
$$X=\frac{(\cos^2x-\sin^2x)\times1}{\cos^2x}$$
$$X=\frac{\cos^2x-\sin^2x}{\cos^2x}$$
$$X=1-\frac{\sin^2x}{\cos^2x}$$
- As $\tan x=\frac{\sin x}{\cos x}$, it follows that $\tan^2x=\frac{\sin^2x}{\cos^2x}$
$$X=1-\tan^2x$$
So, $$\frac{\cos^4x-\sin^4x}{\cos^2x}=1-\tan^2x$$
The equation is an identity as a result.