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If $A$ and $B$ are positive acute angles satisfying $3 \cos ^{2} A+2 \cos ^{2} B=4$ and $\frac{3 \sin A}{\sin B}=\frac{2 \cos B}{\cos A}$
Then the value of $\mathrm{A}+2 \mathrm{~B}$ is equal to :
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Then the value of $\mathrm{A}+2 \mathrm{~B}$ is equal to :
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Verified Answer
The correct answer is:
$\frac{\pi}{2}$
Given, $3 \cos ^{2} \mathrm{~A}+2 \cos ^{2} \mathrm{~B}=4$
$\Rightarrow 2 \cos ^{2} \mathrm{~B}-1=4-3 \cos ^{2} \mathrm{~A}-1$
$\Rightarrow \cos 2 \mathrm{~B}=3\left(1-\cos ^{2} \mathrm{~A}\right)=3 \sin ^{2} \mathrm{~A} \ldots(1)$
and $2 \cos \mathrm{B} \sin \mathrm{B}=3 \sin \mathrm{A} \cos \mathrm{A}$
$\sin 2 \mathrm{~B}=3 \sin \mathrm{A} \cos \mathrm{A}$
Now, $\cos (\mathrm{A}+2 \mathrm{~B})=\cos \mathrm{A} \cos 2 \mathrm{~B}-\sin \mathrm{A}$
$\sin 2 \mathrm{~B}$
$=\cos A\left(3 \sin ^{2} A\right)-\sin A(3 \sin A \cos A)=0$
[using eqs. (1) and (2)] $\Rightarrow A+2 B=\frac{\pi}{2}$
$\Rightarrow 2 \cos ^{2} \mathrm{~B}-1=4-3 \cos ^{2} \mathrm{~A}-1$
$\Rightarrow \cos 2 \mathrm{~B}=3\left(1-\cos ^{2} \mathrm{~A}\right)=3 \sin ^{2} \mathrm{~A} \ldots(1)$
and $2 \cos \mathrm{B} \sin \mathrm{B}=3 \sin \mathrm{A} \cos \mathrm{A}$
$\sin 2 \mathrm{~B}=3 \sin \mathrm{A} \cos \mathrm{A}$
Now, $\cos (\mathrm{A}+2 \mathrm{~B})=\cos \mathrm{A} \cos 2 \mathrm{~B}-\sin \mathrm{A}$
$\sin 2 \mathrm{~B}$
$=\cos A\left(3 \sin ^{2} A\right)-\sin A(3 \sin A \cos A)=0$
[using eqs. (1) and (2)] $\Rightarrow A+2 B=\frac{\pi}{2}$
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