Question

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 $A +2 B$ is equal to :

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{2}$
• C. $\frac{\pi}{3}$
• D. $\frac{\pi}{4}$

Answer 1

Answer: (B)

Given, $3 \cos ^{2} A+2 \cos ^{2} B=4$
$\Rightarrow 2 \cos ^{2} B-1=4-3 \cos ^{2} A-1$
$\Rightarrow \cos 2 B=3\left(1-\cos ^{2} A\right)=3 \sin ^{2} A \ldots(1)$
and $2 \cos B \sin B=3 \sin A \cos A \sin 2 B=3 \sin A \cos A \ldots(2)$
Now, $\cos (A+2 B)=\cos A \cos 2 B-\sin A \sin 2 B$
$=\cos A\left(3 \sin ^{2} A\right)-\sin A(3 \sin A \cos A)=0$
[using eqs. (1) and (2)]
$\Rightarrow A+2 B=\frac{\pi}{2}$