If A = [ cases 0 & - ((θ)/(2)) ((θ)/(2)) & 0 cases ] and ( I _ 2 +...

A = \left[\begin{cases}0 & - tan\dfrac{\theta}{2} \\ tan\dfrac{\theta}{2} & 0\end{cases}\right]

⇒ I + A = \left[\begin{cases}1 & - tan\dfrac{\theta}{2} \\ tan\dfrac{\theta}{2} & 1\end{cases}\right]

⇒ I - A = \left[\begin{cases}1 & tan\dfrac{\theta}{2} \\ - tan\dfrac{\theta}{2} & 1\end{cases}\right]\left\{∴ | I - A | = \sec^{2}\theta / 2\right\}

⇒ {\left(I - A\right)}^{- 1} = \dfrac{1}{\sec^{2}\dfrac{\theta}{2}}\left[\begin{cases}1 & - tan\dfrac{\theta}{2} \\ tan\dfrac{\theta}{2} & 1\end{cases}\right]

⇒ \left(I + A\right){\left(I - A\right)}^{- 1} = \dfrac{1}{\sec^{2}\dfrac{\theta}{2}}\left[\begin{cases}1 & - tan\dfrac{\theta}{2} \\ tan\dfrac{\theta}{2} & 1\end{cases}\right]\left[\begin{cases}1 & - tan\dfrac{\theta}{2} \\ tan\dfrac{\theta}{2} & 1\end{cases}\right]

= \dfrac{1}{\sec^{2}\dfrac{\theta}{2}}\left[\begin{cases}1 - \tan^{2}\dfrac{\theta}{2} & - 2tan\dfrac{\theta}{2} \\ 2tan\dfrac{\theta}{2} & 1 - \tan^{2}\dfrac{\theta}{2}\end{cases}\right]

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

$b = \dfrac{2tan\dfrac{\theta}{2}}{\sec^{2}\dfrac{\theta}{2}}$

$∴ {a}^{2} + {b}^{2} = 1$

$∴ 13\left({a}^{2} + {b}^{2}\right) = 13$