Question

If $AandB$ be acute positive angles satisfying $3si{n}^{2}A+2si{n}^{2}B=1and3sin2A-2sin2B=0$, then

• A. $B=\frac{\pi }{4}-\frac{A}{2}$
• B. $A=\frac{\pi }{4}-2B$
• C. $B=\frac{\pi }{2}-\frac{A}{4}$
• D. $A=\frac{\pi }{4}-\frac{B}{2}$

Answer 1

The correct option is A $B=\frac{\pi }{4}-\frac{A}{2}$

From the given relations, we have
$\sin 2B=\frac{3}{2}\sin 2A$ and $3{\sin }^{2}A=1-2{\sin }^{2}B=\cos 2B$
So that
$\cos (A+2B)=\cos A\cos 2B-\sin A\sin 2B$
$=\cos A.3{\sin }^{2}A-\frac{3}{2}\sin A\sin 2A$
$=3\cos A{\sin }^{2}A-3{\sin }^{2}A\cos A=0$
$\Rightarrow A+2B=\frac{\pi }{2}$

$\therefore B=\frac{\pi }{4}-\frac{A}{2}$