Question

If $A$ and $B$ are two positive acute angles satisfying the equations $4-3{\cos }^{2}A=2{\cos }^{2}B$ and $\cos (A+2B)=0$, then the value of $\frac{3\sin A}{2\cos B}$ is

• A. $\frac{3\cos A}{\sin B}$
• B. $\frac{\cos B}{\sin A}$
• C. $\frac{2\cos B}{\sin B}$
• D. $\frac{\sin B}{\cos A}$

Answer 1

The correct option is D $\frac{\sin B}{\cos A}$

$4-3{\cos }^{2}A=2{\cos }^{2}B$
$\Rightarrow 3-3{\cos }^{2}A+1=2{\cos }^{2}B$
$\Rightarrow 3(1-{\cos }^{2}A)=2{\cos }^{2}B-1$
$\Rightarrow 3{\sin }^{2}A=\cos 2B\cdots \left(1\right)$

$\cos (A+2B)=\cos A\cos 2B-\sin A\sin 2B=0$
$\Rightarrow \cos A\cos 2B=\sin A\sin 2B$
From eqn $\left(1\right)$
$3\cos A{\sin }^{2}A=\sin A\sin 2B$
$\Rightarrow 3\cos A\sin A=\sin 2B$
$\Rightarrow 3\cos A\sin A=2\sin B\cos B$
$\Rightarrow \frac{3\sin A}{2\cos B}=\frac{\sin B}{\cos A}$