Question

If $\alpha$ and $\beta$ are roots of equation $\frac{3}{4}\sin \left(\frac{\theta }{9}\right)={\sin }^3\theta +3{\sin }^3\left(\frac{\theta }{3}\right)+9{\sin }^3\left(\frac{\theta }{9}\right)+\frac{1}{4\sqrt{2}}$ for $0<\theta <\frac{\pi }{2}$ then $\tan \alpha +\tan \beta$ is equal to

• A. $2+\sqrt{3}$
• B. $3+\sqrt{3}$
• C. $3-\sqrt{3}$
• D. $2-\sqrt{3}$

Answer 1

The correct option is C $3-\sqrt{3}$

$\frac{1}{4}\sin 3\theta =\frac{3}{4}\sin \left(\frac{\theta }{9}\right)-{\sin }^3\theta -3{\sin }^3\left(\frac{\theta }{3}\right)-9{\sin }^3\left(\frac{\theta }{9}\right)$

$\frac{3}{4}\sin \left(\frac{\theta }{9}\right)={\sin }^3\theta +3{\sin }^3\left(\frac{\theta }{3}\right)+9{\sin }^3\left(\frac{\theta }{9}\right)+\frac{1}{4\sqrt{2}}$

$⟹\frac{3}{4}\sin \left(\frac{\theta }{9}\right)-{\sin }^3\theta -3{\sin }^3\left(\frac{\theta }{3}\right)-9{\sin }^3\left(\frac{\theta }{9}\right)=\frac{1}{4\sqrt{2}}$

$⟹\frac{1}{4}\sin 3\theta =\frac{1}{4\sqrt{2}}$

$⟹\sin 3\theta =\frac{1}{\sqrt{2}}⟹3\theta =\frac{\pi }{4},\frac{3\pi }{4}⟹\theta =\frac{\pi }{12},\frac{\pi }{4}$

So $\alpha =\frac{\pi }{12},\beta =\frac{\pi }{4}$

$\tan \alpha +\tan \beta =\tan \frac{\pi }{12}+\tan \frac{\pi }{4}=2-\sqrt{3}+1=3-\sqrt{3}$