Q.If $A = \begin{bmatrix} 0 & -\tan\dfrac{\alpha}{2} \\ \tan\dfrac{\alpha}{2} & 0 \end{bmatrix}$, then prove that $(I + A) = (I - A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$.

express both sides with t = tan(α/2) using the half-angle formulas for sinα, cosα

Let $t=\tan\dfrac{\alpha}{2}$, so $A=\begin{bmatrix}0&-t\\t&0\end{bmatrix}$, $I+A=\begin{bmatrix}1&-t\\t&1\end{bmatrix}$, $I-A=\begin{bmatrix}1&t\\-t&1\end{bmatrix}$.

Recall the half-angle formulas: $\cos\alpha=\dfrac{1-t^2}{1+t^2}$, $\sin\alpha=\dfrac{2t}{1+t^2}$.

Compute $(I-A)\begin{bmatrix}\cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha\end{bmatrix}$:

$\begin{bmatrix}1&t\\-t&1\end{bmatrix}\begin{bmatrix}\cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha\end{bmatrix}=\begin{bmatrix}\cos\alpha+t\sin\alpha&-\sin\alpha+t\cos\alpha\\-t\cos\alpha+\sin\alpha&t\sin\alpha+\cos\alpha\end{bmatrix}$

Entry $(1,1)$: $\cos\alpha+t\sin\alpha=\dfrac{1-t^2}{1+t^2}+\dfrac{2t^2}{1+t^2}=\dfrac{1+t^2}{1+t^2}=1$

Entry $(1,2)$: $-\sin\alpha+t\cos\alpha=\dfrac{-2t+t(1-t^2)}{1+t^2}=\dfrac{-t-t^3}{1+t^2}=-t$

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