Question

Show that, $\left[\begin{array}{cc}1& -\tan \frac{\theta }{2}\\ \tan \frac{\theta }{2}& 1\end{array}\right]{\left[\begin{array}{cc}1& \tan \frac{\theta }{2}\\ -\tan \frac{\theta }{2}& 1\end{array}\right]}^{-1}=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right].$

Answer 1

As ${\left[\begin{array}{cc}1& \tan \frac{\theta }{2}\\ -\tan \frac{\theta }{2}& 1\end{array}\right]}^{-1}=\frac{1}{1+{\tan }^2\frac{\theta }{2}}\left[\begin{array}{cc}1& -\tan \frac{\theta }{2}\\ \tan \frac{\theta }{2}& 1\end{array}\right]$

Hence

$\left[\begin{array}{cc}1& -\tan \frac{\theta }{2}\\ \tan \frac{\theta }{2}& 1\end{array}\right]{\left[\begin{array}{cc}1& \tan \frac{\theta }{2}\\ -\tan \frac{\theta }{2}& 1\end{array}\right]}^{-1}$

$=\frac{1}{1+{\tan }^2\frac{\theta }{2}}\left[\begin{array}{cc}1& -\tan \frac{\theta }{2}\\ \tan \frac{\theta }{2}& 1\end{array}\right]\left[\begin{array}{cc}1& -\tan \frac{\theta }{2}\\ \tan \frac{\theta }{2}& 1\end{array}\right]$

$=\frac{1}{{\sec }^2\frac{\theta }{2}}\left[\begin{array}{cc}1-{\tan }^2\frac{\theta }{2}& -2\tan \frac{\theta }{2}\\ 2\tan \frac{\theta }{2}& 1-{\tan }^2\frac{\theta }{2}\end{array}\right]$

$=\frac{1}{{\sec }^2\frac{\theta }{2}}\left[\begin{array}{cc}\frac{{\sin }^2\frac{\theta }{2}-{\cos }^2\frac{\theta }{2}}{{\cos }^2\frac{\theta }{2}}& -2\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}\\ 2\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}& \frac{{\sin }^2\frac{\theta }{2}-{\cos }^2\frac{\theta }{2}}{{\cos }^2\frac{\theta }{2}}\end{array}\right]$

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