Question

If $\alpha ,\beta ,\gamma ,\delta$, are the solution of the equation $\tan (\theta +\frac{\pi }{4})=3\tan 3\theta$, no two of which have equal tangents.

The value of $\tan \alpha +\tan \beta +\tan \gamma +\tan \delta$ is

• A. $\frac{1}{3}$
• B. $\frac{8}{3}$
• C. $-\frac{8}{3}$
• D. $0$

Answer 1

Answer: (D)

$0$

Given:

$\Rightarrow$$\tan (\theta +\frac{\pi }{4})=3\tan 3\theta$

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

$\Rightarrow {\tan }^4\theta -2{\tan }^2\theta +\frac{8}{3}\tan \theta -\frac{1}{3}=0$

$\Rightarrow {\tan }^4\theta +0{\tan }^3\theta -2{\tan }^2\theta +\frac{8}{3}\tan \theta -\frac{1}{3}=0$

Therefore sum of roots is,

$\tan \alpha +\tan \beta +\tan \gamma +\tan \delta =0$