Question

If A and B be acute +ive angles satisfying the equalities $3{\sin }^{2}A+2{\sin }^{2}B=1$ and $3\sin 2A-2\sin 2B=0$.

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

Answer 1

$3si{n}^{2}A=1-2si{n}^{2}B=cos2B$

Also $6\sin A\cos A=2\sin 2B$

Dividing , We get

$2\cot A=2\tan 2B$

or$\tan 2B=\cot A=\tan (\frac{\pi }{2}-A)$

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