Question

If $\alpha ,\beta ,\gamma ,\delta$ are the solutions of the equation $\tan (\theta +\frac{\pi }{4})=3\tan 3\theta$ no two of which have equal tangents, then prove that $\tan \alpha +\tan \beta +\tan \gamma +\tan \delta =0$.

Answer 1

Using $\tan (\theta +\frac{\pi }{4})=\frac{1+\tan \theta }{1-\tan \theta }$ and $3\tan 3\theta =\frac{3(3\tan \theta -{\tan }^3\theta )}{1-3{\tan }^2\theta }$
The given equation become
$3{\tan }^4\theta -6{\tan }^2\theta +8\tan \theta -1=0$
If $\tan \alpha ,\tan \beta ,\tan \gamma$ and $\tan \delta$ are the roots of this equation, then the sum of there roots $\tan \alpha +\tan \beta +\tan \gamma +\tan \delta$ is equal to zero