If $\left|\begin{array}{ccc}\cos (A+B) & -\sin (A+B) & \cos (2 B) \\ \sin A & \cos A & \sin B \\ -\cos A &…

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 the value of $\mathrm{B}$ is
  1. $\mathrm{n} \pi, \mathrm{n} \in \mathbb{Z}$
  2. $(2 \mathrm{n}+1) \frac{\pi}{2}, \mathrm{n} \in \mathrm{Z}$
  3. $(2 n+1) \frac{\pi}{4}, n \in \mathbb{Z}$
  4. $2 \mathrm{n} \frac{\pi}{3}, \mathrm{n} \in \mathbb{Z}$

Solution

$\begin{aligned} & \left|\begin{array}{ccc}\cos (A+B) & -\sin (A+B) & \cos (2 B) \\ \sin A & \cos A & \sin B \\ -\cos A & \sin A & \cos B\end{array}\right|=0 \\ & \therefore \quad \cos (\mathrm{A}+\mathrm{B})[(\cos \mathrm{A} \cos \mathrm{B}-\sin \mathrm{A} \sin \mathrm{B})] \\ & +\sin (A+B)[\sin A \cos B+\sin B \cos A] \\ & +\cos 2 \mathrm{~B}\left[\sin ^2 \mathrm{~A}+\cos ^2 \mathrm{~A}\right]=0 \\ & \therefore \quad \cos (\mathrm{A}+\mathrm{B}) \cdot \cos (\mathrm{A}+\mathrm{B}) \\ & +\sin (A+B) \cdot \sin (A+B)+\cos 2 B=0 \\ & \cos ^2(A+B)+\sin ^2(A+B)+\cos 2 B=0 \\ & 1+\cos 2 B=0 \\ & 2 \cos ^2 \mathrm{~B}=0 \\ & \therefore \quad \cos \mathrm{B}=0 \\ & \therefore \quad B=(2 n+1) \frac{\pi}{2} \text { for }(n \in Z) \\ & \end{aligned}$

Asked in: MHT CET 2023 (09 May Shift 2)

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