Answer
$$\sec^4 x-\sec^2 x=\tan^4 x+\tan^2 x$$
The equation is verified to be an identity.
Work Step by Step
$$\sec^4 x-\sec^2 x=\tan^4 x+\tan^2 x$$
2 sides look equally complex, so any choice of simplification is arbitrary. Here I would deal with the left side first, using $\sec x=\frac{1}{\cos x}$ $$\sec^4 x-\sec^2 x$$ $$=\frac{1}{\cos^4 x}-\frac{1}{\cos^2 x}$$ $$=\frac{1-\cos^2 x}{\cos^4 x}$$
As we already know, $\sin^2 x+\cos^2 x=1$. But that also means $1-\cos^2 x=\sin^2 x$. Therefore, $$=\frac{\sin^2 x}{\cos^4 x}$$
On the right side, again we simplify, using $\tan x=\frac{\sin x}{\cos x}$ $$\tan^4 x+\tan^2 x$$ $$=\frac{\sin^4 x}{\cos^4 x}+\frac{\sin^2 x}{\cos^2 x}$$ $$=\frac{\sin^4 x}{\cos^4 x}+\frac{\sin^2 x\cos^2 x}{\cos^4 x}$$ $$=\frac{\sin^4 x+\sin^2 x\cos^2 x}{\cos^4 x}$$ $$=\frac{\sin^2 x(\sin^2 x+\cos^2 x)}{\cos^4 x}$$ $$=\frac{\sin^2 x}{\cos^4 x}$$ (for $\sin^2 x+\cos^2 x=1$)
Both sides equal $\frac{\sin^2 x}{\cos^4 x}$, so they are equal.
$$\sec^4 x-\sec^2 x=\tan^4 x+\tan^2 x$$
The equation is thus verified to be an identity.