Question

If tan $\theta$. tan $({120}^0-\theta )$. tan $({120}^0+\theta )=\frac{1}{\sqrt{3}}$, then $\theta =$

• A. $\frac{n\pi }{3}+\frac{\pi }{18},nϵZ$
• B. $\frac{n\pi }{3}+\frac{\pi }{12},nϵZ$
• C. $\frac{n\pi }{12}+\frac{\pi }{12},nϵZ$
• D. $\frac{n\pi }{3}+\frac{\pi }{6},nϵZ$

Answer 1

The correct option is A $\frac{n\pi }{3}+\frac{\pi }{18},nϵZ$

As we know that

$\sin \theta \sin ({120}^{\circ }-\theta )\sin ({120}^{\circ }+\theta )=\frac{1}{4}\sin 3\theta$

$\cos \theta \cos ({120}^{\circ }-\theta )\cos ({120}^{\circ }+\theta )=\frac{1}{4}\cos 3\theta$

$\tan \theta \tan ({120}^{\circ }-\theta )\tan ({120}^{\circ }+\theta )=\frac{\sin \theta }{\cos \theta }\times \frac{\sin ({120}^{\circ }-\theta )}{\cos ({120}^{\circ }-\theta )}\times \frac{\sin ({120}^{\circ }+\theta )}{\cos ({120}^{\circ }+\theta )}$

$=\frac{\sin \theta \sin ({120}^{\circ }-\theta )\sin ({120}^{\circ }+\theta )}{\cos \theta \cos ({120}^{\circ }-\theta )\cos ({120}^{\circ }+\theta )}$

$=\frac{\frac{1}{4}\sin 3\theta }{\frac{1}{4}\cos 3\theta }=\tan 3\theta$

Given $\tan \theta \tan ({120}^{\circ }-\theta )\tan ({120}^{\circ }+\theta )=\frac{1}{\sqrt{3}}$

$⟹\tan 3\theta =\frac{1}{\sqrt{3}}$

$⟹3\theta =n\pi +\frac{\pi }{6}$

$⟹\theta =\frac{n\pi }{3}+\frac{\pi }{18}$