If 2x = sec A and

(a) $\frac{1}{2}$

Given: $2 x=\sec A$ and $\frac{2}{x}=\tan A$

Also, we can deduce that $\mathrm{x}=\frac{\sec A}{2}$ and $\frac{1}{x}=\frac{\tan A}{2}$.

So, substituting the values of $x$ and $\frac{1}{x}$ in the given expression, we get:

$2\left(x^{2}-\frac{1}{x^{2}}\right)=2\left(\left(\frac{\sec A}{2}\right)^{2}-\left(\frac{\tan A}{2}\right)^{2}\right)$

$=2\left(\left(\frac{\sec ^{2} A}{4}\right)-\left(\frac{\tan ^{2} A}{4}\right)\right)$

$=\frac{2}{4}\left(\sec ^{2} A-\tan ^{2} A\right)$

$=\frac{1}{2} \quad$ [By using the identity: $\left(\sec ^{2} \theta-\tan ^{2} \theta=1\right)$ ]