A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{6}\]
Correct Answer: B
Solution :
[b] \[3{{\sin }^{2}}A+2{{\sin }^{2}}B=1\] Or \[3{{\sin }^{2}}A=\cos 2B\] Also \[3\sin 2A-2\sin 2B=0\] Or \[\sin 2B=\frac{3}{2}\sin 2A\] Now, \[\cos (A+2B)=cosAcos2B-sinAsin2B\] \[=\cos A3{{\sin }^{2}}A-\operatorname{sinA}\frac{3}{2}\sin 2A\] \[=3{{\sin }^{2}}A\cos A=3{{\sin }^{2}}A\operatorname{cosA}=0\] \[\therefore \,A+2B=\frac{\pi }{2}\]
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