Question

If $A$ and $B$ are acute +ove angles satisfying the equation $3{\sin }^{2}A+2{\sin }^{2}B=1$ and $3\sin 2A-2\sin 2B=0$, then prove that $A+2B=\pi /2$

Answer 1

$3{\sin }^{2}A=1-2{\sin }^{2}B=\cos 2B$ and $\sin 2B=\frac{3}{2}\sin 2A$
Now, $\cos (A+2B)=\cos A\cos 2B-\sin A\sin 2B=3\cos A{\sin }^{2}A-\sin A.\frac{3}{2}\sin 2A=3\cos A{\sin }^{2}A-3\cos A{\sin }^{2}A=0$
$\therefore A+2B={90}^o=\frac{\pi }{2}$