Question

If $0<A,B<\frac{\pi }{2}$ satisfying the equation $3{\sin }^{2}A+2.{\sin }^{2}B=1$ and $2\sin 2A-2\sin 2B=0$ then $A+2B=$

• A. $0$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{3}$

Answer 1

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

$3si{n}^{2}A+2si{n}^{2}B=1$

$3sin2A-2sin2B=0$

$\therefore sin2B=\frac{3}{2}sin2A$ _______(1)

& $3si{n}^{2}A=1-2si{n}^{2}B=co{s}^{2}B$

As $cos(A+2B)=cosAcos2B-sinAsin2B$

$=3cosAsi{n}^{2}A-\frac{3}{2}sinAsin2A$

$=3cosAsi{n}^{2}A-3si{n}^{2}AcosA=0$

$\Rightarrow A+2B=\frac{\pi }{2}$ or $\frac{3\pi }{2}$

Give $0<A<\pi /2$ & $0<B<\pi /2$

$\Rightarrow 0<A+2B<\pi +\frac{\pi }{2}$ $\therefore A+2B=\pi /2$