If \( A=\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right] \) and \( A+A^{T}=I \),
where \( \mathrm{I} \) is the unit matrix of \( 2 \times 2 \& \mathrm{~A}^{\mathrm{T}} \) is the transpose of \( \mathrm{A} \), then the value of \( \theta \) is equal to

Given that, \(\mathrm{A}=\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right]\)
So, \(\mathrm{A}^{\mathrm{T}}=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]\)
Now, \(\mathrm{A}+\mathrm{A}^{\mathrm{T}}=\left[\begin{array}{ccc}2 \cos 2 \theta & 0 \\ 0 & 2 \cos 2 \theta\end{array}\right]\)
\(=2 \cos 2 \theta\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]\)
Given that, \(\mathrm{A}+\mathrm{A}^{\mathrm{T}}=\mathrm{I}\) so
\(\Rightarrow(2 \cos 2 \theta) I=I\)
\(\Rightarrow 2 \cos 2 \theta=1 \Rightarrow \cos 2 \theta=\frac{1}{2}\)
\(\Rightarrow 2 \theta=\frac{\pi}{3} \Rightarrow \theta=\frac{\pi}{6}\)