$\begin{aligned} & \mathrm{P}=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \\ & \because \mathrm{P}^{\mathrm{T}} \mathrm{P}=\mathrm{I} \\ & \mathrm{~B}=\mathrm{PAPT} \end{aligned}$
Pre multiply by $\mathrm{P}^{\mathrm{T}}$ ( Given)
$\mathrm{P}^{\mathrm{T}} \mathrm{~B}=\mathrm{P}^{\mathrm{T}} \mathrm{PA} \mathrm{P}^{\mathrm{T}}=\mathrm{AP}^{\mathrm{T}}$
Now post multiply by P
$\mathrm{P}^{\mathrm{T}} \mathrm{BP}=\mathrm{AP}^{\mathrm{T}} \mathrm{P}=\mathrm{A}$
$\mathrm{A}^2=\mathrm{P}^{\mathrm{T}} \mathrm{~B}^2 \mathrm{P}$
Similarly $A^{10}=P^T B^{10} P=C$
$\begin{aligned} & A=\left[\begin{array}{cc} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{array}\right] \text { (Given) } \\ & \Rightarrow A^2=\left[\begin{array}{cc} \frac{1}{2} & -\sqrt{2}-2 \\ 0 & 1 \end{array}\right] \end{aligned}$
Similarly check $\mathrm{A}^3$ and so on since $\mathrm{C}=\mathrm{A}^{10}$
$\Rightarrow$ Sum of diagonal elements of C is $\left(\frac{1}{\sqrt{2}}\right)^{10}+1$
$\begin{aligned} & =\frac{1}{32}+1=\frac{33}{32}=\frac{\mathrm{m}}{\mathrm{n}} \\ & \mathrm{~g} \mathrm{~cd}(\mathrm{~m}, \mathrm{n})=1 \text { (Given) } \\ & \Rightarrow \mathrm{m}+\mathrm{n}=65 \end{aligned}$