If $A=\begin{bmatrix} \cos^{2}\alpha & \cos\alpha\sin\alpha\\ \cos\alpha\sin\alpha & \sin^{2}\alpha\\ \end{bmatrix}$ and $B=\begin{bmatrix} \cos^{2}\beta & \cos\beta\sin\beta\\ \cos\beta\sin\beta & \sin^{2}\beta\\ \end{bmatrix}$ then if $AB=0$, $\alpha-\beta$ is a multiple of

$\pi/3$ $\pi/8$ $\pi/4$ $\pi/2$

$\pi/2$

: $AB=\cos(\alpha-\beta)\begin{bmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta\\ \sin\alpha\cos\beta & \sin\alpha\sin\beta\\ \end{bmatrix}$. Hence $AB=0$ gives $\cos(\alpha-\beta)=0$. So $\alpha-\beta$ is a multiple of $\pi/2$.