Question

If $\tan \theta +\tan (\theta +\frac{\pi }{3})+\tan (\theta -\frac{\pi }{3})=k\tan 3\theta$ then k is equal to

• A. $1$
• B. $3$
• C. $\frac{1}{3}$
• D. none of these

Answer 1

Answer: (B)

$3$

given, $\tan \theta +\tan (\theta +\frac{\pi }{3})+\tan (\theta -\frac{\pi }{3})$

$=\tan \theta +\frac{\tan \theta +\sqrt{3}}{1-\sqrt{3}\tan \theta }+\frac{\tan \theta -\sqrt{3}}{1+\sqrt{3}\tan \theta }$

$=\frac{\tan \theta (1-3{\tan }^2\theta )+(\tan \theta +\sqrt{3})(1+\sqrt{3}\tan \theta )+(\tan \theta -\sqrt{3})(1-\sqrt{3}tan\theta )}{(1-3{\tan }^2\theta )}$

$=\frac{\tan \theta -3ta{n}^3\theta +\tan \theta +\sqrt{3}{\tan }^2\theta +\sqrt{3}+3\tan \theta +\tan \theta -\sqrt{3}{\tan }^2\theta -\sqrt{3}+3\tan \theta }{(1-3{\tan }^2\theta )}$

$=\frac{9\tan \theta -3{\tan }^2\theta }{1-3{\tan }^2\theta }=3\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }=3\tan 3\theta$