Proof from Matrix I have tried upto where I could. please anyone help me to complete my proof..

Help much appreciated 

I therefore ends your calculations:
$\left(
\begin{array}{cc}
1 & \frac{\sin \frac \alpha 2}{\cos \frac \alpha 2} \\
-\frac{\sin \frac \alpha 2}{\cos \frac \alpha 2} & 1
\end{array}
\right) \left(
\begin{array}{cc}
\cos \alpha  & -\sin \alpha  \\
\sin \alpha  & \cos \alpha
\end{array}
\right) =\left(
\begin{array}{cc}
\cos \alpha +\frac{\sin \frac 12\alpha }{\cos \frac 12\alpha }\sin \alpha  &
-\sin \alpha +\frac{\sin \frac 12\alpha }{\cos \frac 12\alpha }\cos \alpha
\\
-\frac{\sin \frac 12\alpha }{\cos \frac 12\alpha }\cos \alpha +\sin \alpha
& \cos \alpha +\frac{\sin \frac 12\alpha }{\cos \frac 12\alpha }\sin \alpha
\end{array}
\right) $ ;
so developing the expressions, in the last matrix,  using trigonometric forms,
$\sin \alpha =2\sin \frac \alpha 2\cos \frac \alpha 2$
$\cos \alpha =2\cos ^2\frac \alpha 2-1$
$\cos ^2\frac \alpha 2+\sin ^2\frac \alpha 2=1$,
the result is obtained as
 $\left(
\begin{array}{cc}
1 & -\frac{\sin \frac 12\alpha }{\cos \frac 12\alpha } \\
\frac{\sin \frac 12\alpha }{\cos \frac 12\alpha } & 1
\end{array}
\right) $
