Answer
$\sec^{4}x-\tan^{4}x=\sec^{2}x+\tan^{2}x$
Work Step by Step
$\sec^{4}x-\tan^{4}x=\sec^{2}x+\tan^{2}x$
On the left side of the equation, replace $\sec^{4}x$ by $\dfrac{1}{\cos^{4}x}$ and $\tan^{4}x$ by $\dfrac{\sin^{4}x}{\cos^{4}x}$:
$\dfrac{1}{\cos^{4}x}-\dfrac{\sin^{4}x}{\cos^{4}x}=\sec^{2}x+\tan^{2}x$
Evaluate the subtraction on the left side:
$\dfrac{1-\sin^{4}x}{\cos^{4}x}=\sec^{2}x+\tan^{2}x$
Factor the numerator on the left:
$\dfrac{(1-\sin^{2}x)(1+\sin^{2}x)}{\cos^{4}x}=\sec^{2}x+\tan^{2}x$
Rewrite the left side as $\dfrac{1-\sin^{2}x}{\cos^{2}x}\cdot\dfrac{1+\sin^{2}x}{\cos^{2}x}$:
$\dfrac{1-\sin^{2}x}{\cos^{2}x}\cdot\dfrac{1+\sin^{2}x}{\cos^{2}x}=\sec^{2}x+\tan^{2}x$
Since $1-\sin^{2}x=\cos^{2}x$, then $\dfrac{1-\sin^{2}x}{\cos^{2}x}=1$:
$(1)\cdot\dfrac{1+\sin^{2}x}{\cos^{2}x}=\sec^{2}x+\tan^{2}x$
$\dfrac{1+\sin^{2}x}{\cos^{2}x}=\sec^{2}x+\tan^{2}x$
Rewrite the left side as $\dfrac{1}{\cos^{2}x}+\dfrac{\sin^{2}x}{\cos^{2}x}$:
$\dfrac{1}{\cos^{2}x}+\dfrac{\sin^{2}x}{\cos^{2}x}=\sec^{2}x+\tan^{2}x$
Since $\dfrac{1}{\cos^{2}x}=\sec^{2}x$ and $\dfrac{\sin^{2}x}{\cos^{2}x}=\tan^{2}x$, the identity is proved
$\sec^{2}x+\tan^{2}x=\sec^{2}x+\tan^{2}x$
