Question

If $A=\left[\begin{array}{cc}0& -\tan \frac{\alpha }{2}\\ \tan \frac{\alpha }{2}& 0\end{array}\right]$ show that $(I+A)=(I-A)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$.

Answer 1

$A=\left[\begin{array}{cc}0& -\tan \frac{\alpha }{2}\\ \tan \frac{\alpha }{2}& 0\end{array}\right],I+A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}0& -\tan \frac{\alpha }{2}\\ \tan \frac{\alpha }{2}& 0\end{array}\right]I+A=\left[\begin{array}{cc}1& -\tan \frac{\alpha }{2}\\ \tan \frac{\alpha }{2}& 1\end{array}\right]⟶\left(1\right)I-A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}0& -\tan \frac{\alpha }{2}\\ \tan \frac{\alpha }{2}& 0\end{array}\right]=I+A=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}1& \tan \frac{\alpha }{2}\\ -\tan \frac{\alpha }{2}& 1\end{array}\right]\therefore (I-A)\cdot \left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]=\left[\begin{array}{cc}1& \tan \frac{\alpha }{2}\\ -\tan \frac{\alpha }{2}& 1\end{array}\right]\cdot \left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]=\left[\begin{array}{cc}\cos \alpha +\sin \alpha \tan \frac{\alpha }{2}& -\sin \alpha +\cos \alpha \cdot \tan \frac{\alpha }{2}\\ -\cos \alpha \cdot \tan \frac{\alpha }{2}+\sin \alpha & \sin \alpha \tan \frac{\alpha }{2}+\cos \alpha \end{array}\right]⟶\left(2\right)\cos \alpha +\sin \alpha \tan \frac{\alpha }{2}\Rightarrow \cos \alpha +\sin \alpha \frac{(1-\cos \alpha )}{\sin \alpha }\Rightarrow \cos \alpha +1-\cos \alpha \Rightarrow 1⟶\left(3\right)-\cos \alpha \cdot \tan \frac{\alpha }{2}+\sin \alpha \Rightarrow -\cos \alpha \cdot \frac{(1-\cos \alpha )}{\sin \alpha }+\sin \alpha \Rightarrow \frac{-\cos \alpha +\cos {}^2\alpha }{\sin \alpha }+\sin \alpha \Rightarrow \frac{-\cos \alpha +\cos {}^2\alpha +\sin {}^2\alpha }{\sin \alpha }\Rightarrow \frac{-\cos \alpha +1}{\sin \alpha }\Rightarrow \tan \frac{\alpha }{2}⟶\left(4\right)\therefore I+A=\left[\begin{array}{cc}1& -\tan \frac{\alpha }{2}\\ \tan \frac{\alpha }{2}& 1\end{array}\right](I-A)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]=\left[\begin{array}{cc}1& -\tan \frac{\alpha }{2}\\ \tan \frac{\alpha }{2}& 1\end{array}\right]\therefore I+A=(I-A)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$

(Hence Proved)