Question

If A and B are acute positive angles satisfying the equations $3{\sin }^{2}A+2{\sin }^{2}B=1$ and $3\sin 2A-2\sin 2B=0$ the $A+2B$ is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{2}$
• C. $\frac{2\pi }{3}$
• D. none

Answer 1

Answer: (B)

$\frac{\pi }{2}$

$3si{n}^{2}A=1-2si{n}^{2}B$
Or
$3si{n}^{2}A=cos2B$.
And
$3sin2A=2sin2B$.
Hence
$\frac{sinA}{2cosA}=\frac{cos2B}{2sin2B}$
Or
$tanA=cot2B$.
Or
$tanA.tan2B=1$.
Now
$tan(A+2B)=\frac{tanA+tan2B}{1-tanA.tanB}$
Now
$1-tanA.tanB=1-1=0$.
Hence

$tan(A+2B)=\frac{tanA+tan2B}{1-tanA.tanB}=\frac{tanA+tan2B}{0}=tan\frac{\pi }{2}$

$\therefore A+2B=\frac{\pi }{2}$.