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If A=[0−tan⁡α/2tan⁡α/20]A=\begin{bmatrix} 0 & -\tan\alpha/2 \\ \tan\alpha/2 & 0 \end{bmatrix}, then the value of [cos⁡α−sin⁡αsin⁡αcos⁡α]\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} for all α∈R\alpha \in \mathbb{R}(set of real numbers), is

(I is identity matrix of order 2)

Solution

✅ Correct Option: 2

Let t=tan⁡(α/2)t=\tan(\alpha/2). Then I+A=[1−tt1]I+A = \begin{bmatrix} 1 & -t \\ t & 1\end{bmatrix} and (I−A)−1=11+t2[1−tt1](I-A)^{-1} = \dfrac{1}{1+t^2}\begin{bmatrix} 1 & -t \\ t & 1\end{bmatrix}.

Multiplying gives 11+t2[1−t2−2t2t1−t2]\dfrac{1}{1+t^2}\begin{bmatrix} 1-t^2 & -2t \\ 2t & 1-t^2\end{bmatrix}.

Using cos⁡α=1−t21+t2\cos\alpha = \dfrac{1-t^2}{1+t^2} and sin⁡α=2t1+t2\sin\alpha = \dfrac{2t}{1+t^2}, this is the given rotation matrix.

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