[1−tanθ2tanθ21] [1tanθ2−tanθ21]−1=

[1−tanθ/2tanθ/21]1sec2θ/2[1−tanθ/2tanθ/21] =cos2θ2[1−tan2θ/2−2tanθ/22tanθ/21−tan2θ/2]= [cosθ−sinθsinθcosθ]

[1−tanθ2tanθ21] [1tanθ2−tanθ21]−1=

[1−tanθ/2tanθ/21]1sec2θ/2[1−tanθ/2tanθ/21] =cos2θ2[1−tan2θ/2−2tanθ/22tanθ/21−tan2θ/2]= [cosθ−sinθsinθcosθ]

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$[\begin{matrix} 1 & - \tan \frac{θ}{2} \\ \tan \frac{θ}{2} & 1 \end{matrix}] {[\begin{matrix} 1 & \tan \frac{θ}{2} \\ - \tan \frac{θ}{2} & 1 \end{matrix}]}^{- 1} =$

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$[\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}]$

$[\begin{matrix} - \cos θ & \sin θ \\ \sin θ & \cos θ \end{matrix}]$

$[\begin{matrix} \sin θ & - \cos θ \\ \cos θ & \sin θ \end{matrix}]$

$[\begin{matrix} - \sin θ & \cos θ \\ \cos θ & - \sin θ \end{matrix}]$

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$[\begin{matrix} 1 & - \tan θ / 2 \\ \tan θ / 2 & 1 \end{matrix}] \frac{1}{\sec^{2} θ / 2} [\begin{matrix} 1 & - \tan θ / 2 \\ \tan θ / 2 & 1 \end{matrix}]$

$= \cos^{2} \frac{θ}{2} [\begin{matrix} 1 - \tan^{2} θ / 2 & - 2 \tan θ / 2 \\ 2 \tan θ / 2 & 1 - \tan^{2} θ / 2 \end{matrix}]$

$= [\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix}]$

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[1−tanθ2tanθ21] [1tanθ2−tanθ21]−1=

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$[\begin{matrix} 1 & - \tan \frac{θ}{2} \\ \tan \frac{θ}{2} & 1 \end{matrix}] {[\begin{matrix} 1 & \tan \frac{θ}{2} \\ - \tan \frac{θ}{2} & 1 \end{matrix}]}^{- 1} =$