Categories > trigonometry > Solution of trigonometric equations > If $\begin{vmatrix} \cos (A + B) & - \sin (A + B) & \cos 2B \\ \sin A & {\cos A} & {\sin B} \\ { - \cos A} & {\sin A} & {\cos B} \end{vmatrix} = 0$ , then B =

trigonometry

Question: If $\begin{vmatrix} \cos (A + B) & - \sin (A + B) & \cos 2B \\ \sin A & {\cos A} & {\sin B} \\ { - \cos A} & {\sin A} & {\cos B} \end{vmatrix} = 0$ , then B =


1) $(2n+1)\frac{\pi }{2}$
2) $n\pi $
3) $(2n+1)\frac{\pi }{2}$
4) $2n\pi $
Solution: Explanation: No Explanation
Solution of trigonometric equations

Rate this question:

★ ★ ★ ★ ★

Average rating: (0 votes)

Previous Question Next Question

Related Questions