Question:
If $A=\left[\begin{array}{cc}0 & -\tan \left(\frac{\theta}{2}\right) \\ \tan \left(\frac{\theta}{2}\right) & 0\end{array}\right]$ and
$\left(\mathrm{I}_{2}+\mathrm{A}\right)\left(\mathrm{I}_{2}-\mathrm{A}\right)^{-1}=\left[\begin{array}{cc}\mathrm{a} & -\mathrm{b} \\ \mathrm{b} & \mathrm{a}\end{array}\right]$, then $13\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)$ is equal to________.
Solution:
$\mathrm{a}^{2}+\mathrm{b}^{2}=\left|\mathrm{I}_{2}+\mathrm{A} \| \mathrm{I}_{2}-\mathrm{A}\right|^{-1}$
$=\sec ^{2} \frac{\theta}{2} \times \cos ^{2} \frac{\theta}{2}=1$
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