Let $$A = \begin{bmatrix}\frac{1}{\sqrt{2}} & -2 \\0 & 1 \end{bmatrix}$$ and $$P = \begin{bmatrix}\cos \theta & -\sin \theta \\\sin \theta & \cos \theta \end{bmatrix}$$ ,$$\theta > 0$$. If $$B = PAP^{T}, C = P^{T}B^{10}P$$ and the sum of the diagonal elements of $$C$$ is $$\frac{m}{n}$$, where $$gcd(m,n)=1,$$ m + n is :

Since $$P$$ is a rotation matrix and thus $$P^T = P^{-1}$$, we have $$B = PAP^T \implies B^{10} = PA^{10}P^T$$. This follows because $$B^2 = (PAP^T)(PAP^T) = PA(P^TP)AP^T = PA^2P^T$$ (since $$P^TP = I$$), and by induction $$B^n = PA^nP^T$$.

Substituting into the expression for $$C$$ gives $$C = P^T B^{10} P = P^T(PA^{10}P^T)P = (P^TP)A^{10}(P^TP) = A^{10}$$, so that $$C = A^{10}$$.

Matrix $$A = \begin{bmatrix} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{bmatrix}$$ is upper triangular, and for such matrices $$A^n$$ remains upper triangular with diagonal elements equal to the $$n$$th powers of the original diagonal elements. Hence the diagonal entries of $$A^{10}$$ are $$\left(\frac{1}{\sqrt{2}}\right)^{10}$$ and $$1^{10}$$.

Therefore, the trace of $$C$$ is $$\text{tr}(C) = \left(\frac{1}{\sqrt{2}}\right)^{10} + 1 = \frac{1}{2^5} + 1 = \frac{1}{32} + 1 = \frac{33}{32}$$, and writing this as $$\frac{m}{n}$$ gives $$\frac{m}{n} = \frac{33}{32}$$ with $$\gcd(33,32)=1$$. It follows that $$m+n = 33+32 = 65$$.

Therefore, the correct answer is Option (3): 65.