If $A=\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{array}\right]$, and $A+A'=I$, then the value of '$\alpha$' is:

$\frac{\pi}{6}$

The correct answer is Option (3) → $\frac{\pi}{6}$

$A+A^T=I$

$⇒\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha\end{array}\right]+\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha\end{array}\right]=I$

$⇒\begin{bmatrix}2\sin α&0\\0&2\sin α\end{bmatrix}=I$

so $2\sin α=1$

$\sin α=\frac{1}{2}$

$α=\frac{\pi}{6}$