A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{4}\]
Correct Answer: B
Solution :
Given, \[3{{\cos }^{2}}A+2{{\cos }^{2}}B=4\] \[\Rightarrow \,\,2{{\cos }^{2}}B-1=4-3{{\cos }^{2}}A-1\] \[\Rightarrow \,\,\cos 2B=3(1-{{\cos }^{2}}A)=3{{\sin }^{2}}A\] ?.(1) and \[2\cos B\,\sin B=3\sin A\,\cos A\] \[\sin 2B=3\sin A\cos A\] ?..(2) Now, \[\cos (A+2B)=\cos A\cos 2B-\sin A\sin 2B\] \[=\cos A(3{{\sin }^{2}}A)-\sin A(3sin\,A\,\,cos\,A)=0\] [using eqs.(1) and (2)] \[\Rightarrow \,\,\,A+2B=\frac{\pi }{2}\]You need to login to perform this action.
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