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If $\left|\begin{array}{ccc}\cos (A+B) & -\sin (A+B) & \cos 2 B \\ \sin A & \cos A & \sin B \\ -\cos A & \sin A & \cos B\end{array}\right|=0$
then $B$ is equal to
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then $B$ is equal to
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Verified Answer
The correct answer is:
$(2 n+1) \frac{\pi}{2}$
We have,
$\left|\begin{array}{ccc}\cos (A+B) & -\sin (A+B) & \cos 2 B \\\sin A & \cos A & \sin B \\-\cos A & \sin A & \cos B\end{array}\right|=0$
$\Rightarrow \quad \cos (A+B)[\cos A \cos B-\sin A \sin B]$ $+\sin (A+B)[\sin A \cos B+\cos A \sin B]$ $+\cos 2 B\left[\sin ^2 A+\cos ^2 A\right]=0$
$\Rightarrow \quad \cos (A+B) \cos (A+B)$ $+\sin (A+B) \sin (A+B)+\cos 2 B=0$
$\Rightarrow \cos ^2(A+B)+\sin ^2(A+B)+\cos 2 B=0$
$\Rightarrow \quad 1+\cos 2 B=0$
$\Rightarrow \quad 1+2 \cos ^2 B-1=0$
$\Rightarrow \quad \cos ^2 B=0 \Rightarrow \cos ^2 B=\cos ^2 \frac{\pi}{2}$
$\Rightarrow \quad B=n \pi \pm \frac{\pi}{2}=\frac{\pi}{2}(2 n \pm 1)$
$\therefore \quad B=\frac{\pi}{2}(2 n+1)$
$\left|\begin{array}{ccc}\cos (A+B) & -\sin (A+B) & \cos 2 B \\\sin A & \cos A & \sin B \\-\cos A & \sin A & \cos B\end{array}\right|=0$
$\Rightarrow \quad \cos (A+B)[\cos A \cos B-\sin A \sin B]$ $+\sin (A+B)[\sin A \cos B+\cos A \sin B]$ $+\cos 2 B\left[\sin ^2 A+\cos ^2 A\right]=0$
$\Rightarrow \quad \cos (A+B) \cos (A+B)$ $+\sin (A+B) \sin (A+B)+\cos 2 B=0$
$\Rightarrow \cos ^2(A+B)+\sin ^2(A+B)+\cos 2 B=0$
$\Rightarrow \quad 1+\cos 2 B=0$
$\Rightarrow \quad 1+2 \cos ^2 B-1=0$
$\Rightarrow \quad \cos ^2 B=0 \Rightarrow \cos ^2 B=\cos ^2 \frac{\pi}{2}$
$\Rightarrow \quad B=n \pi \pm \frac{\pi}{2}=\frac{\pi}{2}(2 n \pm 1)$
$\therefore \quad B=\frac{\pi}{2}(2 n+1)$
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