Correct Answer: $\tan^{2}\alpha$

Solution : Given: $\cos^{2}\alpha–\sin^{2}\alpha=\tan^{2}\beta$
We know that: $1+\tan^{2}\alpha=\sec^{2}\alpha$
So, $\cos^{2}\alpha-\sin^{2}\alpha=\sec^{2}\beta-1$
⇒ $\cos^{2}\alpha-\sin^{2}\alpha+1=\sec^{2}\beta$
⇒ $\cos^{2}\beta=\frac{1}{\cos^{2}\alpha–\sin^{2}\alpha+(\sin^{2}\alpha+\cos^{2}\alpha)}$
⇒ $\cos^{2}\beta=\frac{1}{2\cos^{2}\alpha}$
Also, $\sin^{2}\beta=1-\cos^{2}\beta=1-\frac{1}{2\cos^{2}\alpha}$
So, $\cos^{2}\beta-\sin^{2}\beta$
$=\frac{1}{2\cos^{2}\alpha}-(1-\frac{1}{2\cos^{2}\alpha})$
$= \frac{1}{\cos^{2}\alpha}-1$
$= \sec^{2}\alpha-1$
$= \tan^{2}\alpha$
Hence, the correct answer is $\tan^{2}\alpha$.