Question

If $A$ and $B$ be acute positive angles satisfying $3{\sin }^{2}A+2{\sin }^{2}B=1$ and $3\sin 2A-2\sin 2B=0$, 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

$\sin 2B=\left(\frac{3}{2}\right)\sin 2A$

and $3{\sin }^{2}A=1-2{\sin }^{2}B$

$\Rightarrow 3{\sin }^{2}A={\cos }^{2}B-{\sin }^{2}B=\cos 2B$

so that

$\tan 2B=\frac{sin2B}{\cos 2B}=\frac{\frac{3}{2}.\sin 2A}{3{\sin }^{2}A}$

$=\frac{2\sin A\cos A}{2{\sin }^{2}A}=\cot A$

$\Rightarrow \tan 2B\tan A=1$

$\Rightarrow 1-\tan 2B\tan A=0$

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