Question
If $$A$$ and $$B$$ are positive acute angles satisfying $$3\,{\cos ^2}A + 2\,{\cos ^2}B = 4{\text{ and }}\frac{{3\sin A}}{{\sin B}} = \frac{{2\cos B}}{{\cos A}}.$$ Then the value of $$A + 2B$$ is equal to :
A.
$$\frac{\pi }{6}$$
B.
$$\frac{\pi }{2}$$
C.
$$\frac{\pi }{3}$$
D.
$$\frac{\pi }{4}$$
Answer :
$$\frac{\pi }{2}$$
Solution :
$$\eqalign{
& {\text{Given}},3\,{\cos ^2}A + 2\,{\cos ^2}B = 4 \cr
& \Rightarrow 2\,{\cos ^2}B - 1 = 4 - 3\,{\cos ^2}A - 1 \cr
& \Rightarrow \cos 2B = 3\left( {1 - {{\cos }^2}A} \right) = 3\,{\sin ^2}A\,\,.....\left( 1 \right) \cr
& {\text{and }}2\cos B\sin B = 3\sin A\cos A \cr
& \sin 2B = 3\sin A\cos A\,\,.....\left( 2 \right) \cr
& {\text{Now}},\cos \left( {A + 2B} \right) = \cos A\cos 2B - \sin A\sin 2B \cr
& = \cos A\left( {3\,{{\sin }^2}A} \right) - \sin A\left( {3\sin A\cos A} \right) = 0\left[ {{\text{using eqs}}{\text{. }}\left( 1 \right){\text{and}}\left( 2 \right)} \right] \cr
& \Rightarrow A + 2B = \frac{\pi }{2} \cr} $$