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# If $A = \begin{bmatrix} 0 & -tan \frac{\alpha}{2} \\ tan\frac{\alpha}{2} & 0 \end{bmatrix}$ and $I$ is the identity matrix of order $2$, show that $I + A = ( I - A ) \begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}$

Toolbox:
• Multiplication of two matrices is defined only if the number of columns of the left matrix is the same as the number of rows of the right matrix.
• If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B:
• $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
• The sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where 1 ≤ i ≤ m and 1 ≤ j ≤ n.
• An identity matrix or unit matrix of size n is the n × n square matrix with ones on the main diagonal and zeros elsewhere. An identity matrix of order 2, $I_{2}= \begin{bmatrix} 1 &0 \\ 0&1 \end{bmatrix}$
• $\cos\alpha= \frac{1-\tan^2\frac{\alpha}{2}}{1+\tan\frac{\alpha}{2}}$ and $\sin \alpha =\frac{2\tan\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}$
Given A, we need to show that $:LHS: I+A = RHS: (I-A)\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}$
$LHS: I+A=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}+\begin{bmatrix} 0 & -tan \frac{\alpha}{2} \\ tan\frac{\alpha}{2} & 0 \end{bmatrix}$
$LHS =\begin{bmatrix} 1 & -tan \frac{\alpha}{2} \\ tan\frac{\alpha}{2} & 1 \end{bmatrix}$
Substituting, $\tan \frac{\alpha}{2} = x$, we get $LHS =\begin{bmatrix} 1 & -x \\ \ x & 1 \end{bmatrix}$
$RHS: (I-A)\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}$
$I - A = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} - \begin{bmatrix} 0 & -tan \frac{\alpha}{2} \\ tan\frac{\alpha}{2} & 0 \end{bmatrix}$
$I-A = \begin{bmatrix} 1 & tan \frac{\alpha}{2} \\ -tan\frac{\alpha}{2} & 1 \end{bmatrix}$
Substituting, $\tan \frac{\alpha}{2} = x$, we get $I - A = \begin{bmatrix} 1 & x \\ -x & 1\end{bmatrix}$
We know that $\cos\alpha= \frac{1-\tan^2\frac{\alpha}{2}}{1+\tan\frac{\alpha}{2}}$ and $\sin \alpha =\frac{2\tan\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}$
Therefore $\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix} = \begin{bmatrix}\frac{1-\tan^2\frac{\alpha}{2}}{1+\tan\frac{\alpha}{2}} & -\frac{2\tan\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}\\\frac{2\tan\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}} & \frac{1-\tan^2\frac{\alpha}{2}}{1+\tan\frac{\alpha}{2}}\end{bmatrix}$
Substituting, $\tan \frac{\alpha}{2} = x, \begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix} =\begin{bmatrix}\frac{1-x^2}{1+x^2} & -\frac{2x}{1+x^2}\\\frac{2x}{1+x^2} & \frac{1-x^2}{1+x^2}\end{bmatrix}$
Therefore, $RHS: (I-A)\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix} = \begin{bmatrix} 1 & x \\ -x & 1\end{bmatrix} \begin{bmatrix}\frac{1-x^2}{1+x^2} & -\frac{2x}{1+x^2}\\\frac{2x}{1+x^2} & \frac{1-x^2}{1+x^2}\end{bmatrix}$
$RHS = \begin{bmatrix}\frac{1-x^2}{1+x^2} +\frac{2x^2}{1+x^2}& -\frac{2x}{1+x^2}+\frac{x(1-x^2)}{1+x^2}\\\frac{-x(1-x^2)}{1+x^2}+\frac{2x}{1+x^2} & \frac{2x^2}{1+x^2}+\frac{1-x^2}{1+x^2}\end{bmatrix}$
$RHS = \begin{bmatrix}\frac{1+x^2}{1+x^2} & \frac{-2x+x-x^3}{1+x^2}\\\frac{-x+x^3+2x}{1+x^2} & \frac{1+x^2}{1+x^2}\end{bmatrix}$
Simplifying, $RHS = \begin{bmatrix} 1 & \frac{-x(1+x^2)}{1+x^2}\\\frac{x(1+x^2)}{1+x^2} & 1\end{bmatrix}$
Therefore, $RHS = \begin {bmatrix} 1 & -x \\ x & 1 \end{bmatrix} = LHS$.