If A=[ … ] and (I_2+A)(I_2-A)^-1=[ … ], then 13(a^2+b^2) is equal to .

$$A=[\begin{matrix}0 & -\mathrm{tan}\frac{\theta }{2} \\ \mathrm{tan}\frac{\theta }{2} & 0\end{matrix}]$$A=[0tan2θ​​−tan2θ​0​]

$$\Rightarrow I+A=[\begin{matrix}1 & -\mathrm{tan}\frac{\theta }{2} \\ \mathrm{tan}\frac{\theta }{2} & 1\end{matrix}]$$⇒I+A=[1tan2θ​​−tan2θ​1​]

$$\Rightarrow I-A=[\begin{matrix}1 & \mathrm{tan}\frac{\theta }{2} \\ -\mathrm{tan}\frac{\theta }{2} & 1\end{matrix}]{\therefore |I-A|={\mathrm{sec}}^{2}\theta /2}$$⇒I−A=[1−tan2θ​​tan2θ​1​]∴∣I−A∣=sec2θ/2

$$\Rightarrow {(I-A)}^{-1}=\frac{1}{{\mathrm{sec}}^{2}\frac{\theta }{2}}[\begin{matrix}1 & -\mathrm{tan}\frac{\theta }{2} \\ \mathrm{tan}\frac{\theta }{2} & 1\end{matrix}]$$⇒(I−A)−1=sec22θ​1​[1tan2θ​​−tan2θ​1​]

$$\Rightarrow (I+A){(I-A)}^{-1}=\frac{1}{{\mathrm{sec}}^{2}\frac{\theta }{2}}[\begin{matrix}1 & -\mathrm{tan}\frac{\theta }{2} \\ \mathrm{tan}\frac{\theta }{2} & 1\end{matrix}][\begin{matrix}1 & -\mathrm{tan}\frac{\theta }{2} \\ \mathrm{tan}\frac{\theta }{2} & 1\end{matrix}]$$⇒(I+A)(I−A)−1=sec22θ​1​[1tan2θ​​−tan2θ​1​][1tan2θ​​−tan2θ​1​]

$$=\frac{1}{{\mathrm{sec}}^{2}\frac{\theta }{2}}[\begin{matrix}1-{\mathrm{tan}}^{2}\frac{\theta }{2} & -2\mathrm{tan}\frac{\theta }{2} \\ 2\mathrm{tan}\frac{\theta }{2} & 1-{\mathrm{tan}}^{2}\frac{\theta }{2}\end{matrix}]$$=sec22θ​1​[1−tan22θ​2tan2θ​​−2tan2θ​1−tan22θ​​]

$$a=\frac{1-{\mathrm{tan}}^{2}\frac{\theta }{2}}{{\mathrm{sec}}^{2}\frac{\theta }{2}}$$a=sec22θ​1−tan22θ​​

$$b=\frac{2\mathrm{tan}\frac{\theta }{2}}{{\mathrm{sec}}^{2}\frac{\theta }{2}}$$b=sec22θ​2tan2θ​​

$$\therefore {a}^{2}+{b}^{2}=1$$∴a2+b2=1

$$\therefore 13({a}^{2}+{b}^{2})=13$$∴13(a2+b2)=13