If $A = \left[ {\begin{array}{cccccccccccccccccccc}0&{ - \tan \cfrac{\alpha }{2}}\\{\tan \cfrac{\alpha }{2}}&0\end{array}} \right]$ and I is the identity matrix of order 2,

then show that I + A = (I$-$ A)$\left[ {\begin{array}{cccccccccccccccccccc}{\cos \alpha }&{ - \sin \alpha }\\{cin\alpha }&{\cos \alpha }\end{array}} \right]$.

.:

We know that,
$\cos \alpha = \cfrac{{1 - {{\tan }^2}\left( {\cfrac{\alpha }{2}} \right)}}{{1 + {{\tan }^2}\left( {\cfrac{\alpha }{2}} \right)}} = \cfrac{{1 - {t^2}}}{{1 + {t^2}}}$,

where $\tan \cfrac{\alpha }{2} = t$

and $\sin \alpha = \cfrac{{2\tan \left( {\cfrac{\alpha }{2}} \right)}}{{1 + {{\tan }^2}\left( {\cfrac{\alpha }{2}} \right)}} = \cfrac{{2t}}{{1 + {t^2}}}$

Now, $I + A = \left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right] + \left[ {\begin{array}{cccccccccccccccccccc}0&{ - t}\\t&0\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&{ - t}\\t&1\end{array}} \right]$

and $I - A = \left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right] - \left[ {\begin{array}{cccccccccccccccccccc}0&{ - t}\\t&0\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&t\\{ - t}&1\end{array}} \right]$
$\therefore$ $(I - A)\left[ {\begin{array}{cccccccccccccccccccc}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&t\\{ - t}&1\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}{\cfrac{{(1 - {t^2})}}{{(1 + {t^2})}}}&{\cfrac{{ - 2t}}{{(1 + {t^2})}}}\\{\cfrac{{2t}}{{(1 + {t^2})}}}&{\cfrac{{(1 - {t^2})}}{{(1 + {t^2})}}}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{\cfrac{{1 - {t^2}}}{{1 + {t^2}}} + \cfrac{{2{t^2}}}{{1 + {t^2}}}}&{\cfrac{{ - 2t}}{{1 + {t^2}}} + \cfrac{{t(1 - {t^2})}}{{1 + {t^2}}}}\\{\cfrac{{ - t(1 - {t^2})}}{{1 + {t^2}}} + \cfrac{{2t}}{{1 + {t^2}}}}&{\cfrac{{2{t^2}}}{{(1 + {t^2})}} + \cfrac{{(1 - {t^2})}}{{(1 + {t^2})}}}\end{array}} \right]$
$= \left[ {\begin{array}{cccccccccccccccccccc}1&{ - t}\\t&1\end{array}} \right] = (I + A)$

Hence, $(I - A)\left[ {\begin{array}{cccccccccccccccccccc}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right] = (I + A).$