Question

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

Answer 1

$\theta -\varphi =\frac{\pi }{4}$

$\left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta \end{array}\right]\left[\begin{array}{cc}{\cos }^2\varphi & \cos \varphi \sin \varphi \\ \cos \varphi \sin \varphi & {\sin }^2\varphi \end{array}\right]$

$=\left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta \end{array}\right]\left[\begin{array}{cc}{\cos }^2\left(\frac{\pi }{2}-\theta \right)& \cos \left(\frac{\pi }{2}-\theta \right)\sin \left(\frac{\pi }{2}-\theta \right)\\ \cos \left(\frac{\pi }{2}-\theta \right)\sin \left(\frac{\pi }{2}-\theta \right)& {\sin }^2\left(\frac{\pi }{2}-\theta \right)\end{array}\right]$

$=\left[\begin{array}{cc}{\cos }^2\theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta \end{array}\right]\left[\begin{array}{cc}{\sin }^2\theta & \sin \theta \cos \theta \\ \sin \theta \cos \theta & {\cos }^2\theta \end{array}\right]$

$=\left[\begin{array}{cc}{\sin }^2\theta {\cos }^2\theta -{\sin }^2\theta {\cos }^2\theta & {\sin }^2\theta {\cos }^2\theta -{\sin }^2\theta {\cos }^2\theta \\ {\sin }^2\theta {\cos }^2\theta -{\sin }^2\theta {\cos }^2\theta & {\sin }^2\theta {\cos }^2\theta -{\sin }^2\theta {\cos }^2\theta \end{array}\right]$

$=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=0$