# Prove the following

Question:

If $A=\left[\begin{array}{lr}0 & -\tan \left(\frac{\theta}{2}\right) \\ \tan \left(\frac{\theta}{2}\right) & 0\end{array}\right]$ and $\left(I_{2}+A\right)\left(I_{2}-A\right)^{-1}$

$=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$, then $13\left(a^{2}+b^{2}\right)$ is equal to

Solution:

$\mathrm{A}=\left[\begin{array}{cc}0 & -\tan \frac{\theta}{2} \\ \tan \frac{\theta}{2} & 0\end{array}\right]$

$\Rightarrow I+A=\left[\begin{array}{cc}1 & -\tan \frac{\theta}{2} \\ \tan \frac{\theta}{2} & 1\end{array}\right]$

$\Rightarrow \mathrm{I}-\mathrm{A}=\left[\begin{array}{ccc}1 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} & 1\end{array}\right]$ $\left\{\therefore|\mathrm{I}-\mathrm{A}|=\sec ^{2} \theta / 2\right\}$

$\Rightarrow(\mathrm{I}-\mathrm{A})^{-1}=\frac{1}{\sec ^{2} \frac{\theta}{2}}\left[\begin{array}{cc}1 & -\tan \frac{\theta}{2} \\ \tan \frac{\theta}{2} & 1\end{array}\right]$

$\Rightarrow(\mathrm{I}+\mathrm{A})(\mathrm{I}-\mathrm{A})^{-1}=\frac{1}{\sec ^{2} \frac{\theta}{2}}\left[\begin{array}{cc}1 & -\tan \frac{\theta}{2} \\ \tan \frac{\theta}{2} & 1\end{array}\right]\left[\begin{array}{cc}1 & -\tan \frac{\theta}{2} \\ \tan \frac{\theta}{2} & 1\end{array}\right]$

$=\frac{1}{\sec ^{2} \frac{\theta}{2}}\left[\begin{array}{cc}1-\tan ^{2} \frac{\theta}{2} & -2 \tan \frac{\theta}{2} \\ 2 \tan \frac{\theta}{2} & 1-\tan ^{2} \frac{\theta}{2}\end{array}\right]$

$a=\frac{1-\tan ^{2} \frac{\theta}{2}}{\sec ^{2} \frac{\theta}{2}}$

$\mathbf{b}=\frac{2 \tan \frac{\theta}{2}}{\sec ^{2} \frac{\theta}{2}}$

$\therefore a^{2}+b^{2}=1$