Question

lf $\tan \theta +\tan (\theta +\frac{\pi }{3})+\tan (\theta +\frac{2\pi }{3})=3$ , then which of the following can be equal to $1$.

• A. $\tan 2\theta$
• B. $\tan 3\theta$
• C. ${\tan }^2\theta$
• D. ${\tan }^3\theta$

Answer 1

Answer: (B)

$\tan 3\theta$

$\tan \theta +\tan (\theta +\frac{\pi }{3})+\tan (\theta +\frac{2\pi }{3})=3$

$=\frac{\sin \theta }{\cos \theta }+\frac{\sin (\theta +\frac{\pi }{3})}{\cos (\theta +\frac{\pi }{3})}+\frac{\sin (\theta +\frac{2\pi }{3})}{\cos (\theta +\frac{2\pi }{3})}=3$

$=\frac{\sin \theta }{\cos \theta }+\frac{\sin (\theta +\frac{\pi }{3})\cos (\theta +\frac{2\pi }{3})+\cos (\theta +\frac{\pi }{3})\sin (\theta +\frac{2\pi }{3})}{\cos (\theta +\frac{\pi }{3})\cos (\theta +\frac{2\pi }{3})}=3$

$=\frac{\sin \theta }{\cos \theta }+\frac{2\sin (2\theta +\pi )}{\cos (2\theta +\pi )+\cos \left(\frac{\pi }{3}\right)}=3$

$=\frac{\sin \theta }{\cos \theta }+\frac{-4\sin 2\theta }{-2\cos 2\theta +1}=3$
$=\frac{-2\cos 2\theta \sin \theta +\sin \theta -4\sin 2\theta \cos \theta }{-2\cos 2\theta \cos \theta +\cos \theta }=3$
$=\frac{-2\sin 3\theta +sin\theta -2\sin 2\theta \cos \theta }{-\cos 3\theta -\cos \theta +\cos \theta }=3$
$=\frac{-2\sin 3\theta +\sin \theta -\sin \theta -\sin 3\theta }{-\cos 3\theta }=3$

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

$⟹\tan 3\theta =1$

$\nRightarrow {\tan }^2\theta =1$

So, the correct answer is option B