Question

If $3{\sin }^2\theta +2{\sin }^2\varphi =1$ and $3\sin 2\theta =2\sin 2\varphi$, $0<\theta <\frac{\pi }{2}$ and $0<\varphi <\frac{\pi }{2}$, then the value of $\theta +2\varphi$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\pi$
• D. None of these

Answer 1

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

Given that, $3{\sin }^2\theta +2{\sin }^2\varphi =1$

$\Rightarrow 3{\sin }^2\theta =1-2{\sin }^2\varphi$
$\Rightarrow 3{\sin }^2\theta =\cos 2\varphi$ ...(i)
and $3\sin \theta \cos \theta =\sin 2\varphi$ ...(ii)
On squaring and adding equations (i) and (ii), we get
$9{\sin }^2\theta ({\sin }^2\theta +{\cos }^2\theta )=1$
$\Rightarrow \sin \theta =\frac{1}{3}$ and $\cos \theta =\frac{2\sqrt{2}}{3}$
$\therefore \cos 2\varphi =3\times \frac{1}{9}=\frac{1}{3}$ and $\sin 2\varphi =\frac{2\sqrt{2}}{3}$
Now, $\cos (\theta +2\varphi )=\cos \theta \cos 2\varphi -\sin \theta \sin 2\varphi$
$=\frac{2\sqrt{2}}{3}\cdot \frac{1}{3}-\frac{1}{3}\cdot \frac{2\sqrt{2}}{3}=0$
and $\theta +2\varphi <\frac{3\pi }{2}$
$\therefore \theta +2\varphi =\frac{\pi }{2}$