Question

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

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

Answer 1

Answer: (B)

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