Question

If $A$ and $B$ are positive acute angle satisfying the equation $3{\cos }^{2}A+2{\cos }^{2}B=4$ and $\frac{3\sin A}{\sin B}=\frac{2\cos B}{\cos A},$ then $A+2B$ is

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

Answer 1

The correct option is A $\frac{\pi }{2}$

$\frac{3\sin A}{\sin B}=\frac{2\cos B}{\cos A}\Rightarrow \frac{3\sin A}{\cos A}=\frac{2\cos B}{\cos A}.\frac{\sin B}{\cos A}$

$\Rightarrow \tan A=\frac{\sin 2B}{3{\cos }^{2}A}=\frac{\sin 2B}{\cos 2B}.\frac{\cos 2B}{3{\cos }^{2}A}=\frac{\tan 2B(2{\cos }^{2}B-1)}{3{\cos }^{2}A}$

$=\frac{\tan 2B(4-3{\cos }^{2}A-1)}{3{\cos }^{2}A}(\because 2{\cos }^{2}B=4-3{\cos }^{2}A)$

$=\tan 2B.{\tan }^{2}A$

$\Rightarrow \tan A.\tan 2B=1\Rightarrow \sin A\sin 2B=\cos A\cos 2B$

$\Rightarrow \cos (A+2B)=0\Rightarrow A+2B=\frac{\pi }{2}$