Question

If $\theta -\varphi =\pi /2$, then show that $\left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta \end{array}\right]\left[\begin{array}{cc}{\cos }^2\phi & \cos \phi \sin \phi \\ \cos \phi \sin \phi & {\sin }^2\phi \end{array}\right]=0$

Answer 1

$\begin{array}{c}\left[\begin{array}{c}{\cos }^2\theta \cos \theta \sin \theta \\ \cos \theta \sin \theta {\sin }^2\theta \end{array}\right]\left[\begin{array}{c}{\cos }^2\theta \cos \theta \sin \theta \\ \cos \theta sin\theta {\sin }^2\theta \end{array}\right]\\ =\left[\begin{array}{c}{\cos }^2\theta \cos \theta \sin \theta \\ \cos \theta \sin \theta {\sin }^2\theta \end{array}\right]\left[\begin{array}{c}{\cos }^2\left(\theta -\frac{\pi }{2}\right)\cos \left(\theta -\frac{\pi }{2}\right)\sin \left(\theta -\frac{\pi }{2}\right)\\ \cos \left(\theta -\frac{\pi }{2}\right)\sin \left(\theta -\frac{\pi }{2}\right){\sin }^2\left(\theta -\frac{\pi }{2}\right)\end{array}\right]\\ =\left[\begin{array}{c}{\cos }^2\theta \sin \theta \cos \theta \\ \sin \theta \cos \theta {\sin }^2\theta \end{array}\right]\left[\begin{array}{c}{\sin }^2\theta -\sin \theta \cos \theta \\ -\sin \theta \cos \theta {\cos }^2\theta \end{array}\right]\\ =\left[\begin{array}{c}si{n}^2\theta {\cos }^2\theta -{\sin }^2\theta {\cos }^2\theta -\sin \theta co{s}^3\theta +\sin \theta {\cos }^3\theta \\ {\sin }^3\theta \cos \theta -{\sin }^2\theta \cos \theta -{\sin }^2\theta \cos \theta +{\sin }^2\theta {\cos }^2\theta \end{array}\right]\\ =\left[\begin{array}{c}00\\ 00\end{array}\right]\end{array}$