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If A = \begin{bmatrix} \cos \theta &i\sin \theta \\ i\sin \theta &\cos \theta \end{bmatrix}\left ( \theta =\frac{\pi }{24} \right ) and A^{5}=\begin{bmatrix} a &b \\ c& d \end{bmatrix}, where i=\sqrt{-1}, then which of the following is not true?
Option: 1 0\leq a^{2}+b^{2}\leq 1
Option: 2 a^{2}-d^{2}=0
Option: 3 a^{2}-c^{2}=1
Option: 4 a^{2}-b^{2}=\frac{1}{2}
 

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A^{2}=\left(\begin{array}{cc} \cos 2 \theta & i \sin 2 \theta \\ \operatorname{isin} 2 \theta & \cos 2 \theta \end{array}\right)

\\\text{Similarly}, A^{5}=\left(\begin{array}{cc}\cos 5 \theta & i \sin 5 \theta \\ i \sin 5 \theta & \cos 5 \theta\end{array}\right)=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)\\ (1) a^{2}+b^{2}=\cos ^{2} 5 \theta-\sin ^{2} 5 \theta=\cos 10 \theta=\cos 75^{\circ}\\ (2) a^{2}-d^{2}=\cos ^{2} 5 \theta-\cos ^{2} 5 \theta=0 \\(3) a^{2}-b^{2}=\cos ^{2} 5 \theta+\sin ^{2} 5 \theta=1\\ (4) a^{2}-c^{2}=\cos ^{2} 5 \theta+\sin ^{2} 5 \theta=1

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himanshu.meshram

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