Question

If \[\alpha ,\beta ,\gamma ,\delta \] are the solution of the equation \[\tan \left( {\theta + \dfrac{\pi }{4}} \right) = 3\tan 3\theta \], no two of which have equal tangents. The value of \[\tan \alpha + \tan \beta + \tan \gamma + \tan \delta \] is
A. \[\dfrac{1}{3}\]
B. \[\dfrac{8}{3}\]
C. \[ - \dfrac{8}{3}\]
D. \[0\]

Answer 1

Hint: First of all, expand the given equation by using trigonometric ratios. Then find the sum of the roots of the equation which will lead us to get the final answer. So, use this concept to reach the solution of the given problem.

Complete step by step solution:
Given that \[\tan \left( {\theta + \dfrac{\pi }{4}} \right) = 3\tan 3\theta \]
We know that \[\tan \left( {a + b} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\] and \[\tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}\]
So, we have
\[
   \Rightarrow \dfrac{{\tan \theta + \tan \dfrac{\pi }{4}}}{{1 - \tan \theta \tan \dfrac{\pi }{4}}} = 3\left( {\dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}} \right) \\
   \Rightarrow \dfrac{{\tan \theta + 1}}{{1 - \tan \theta }} = \dfrac{{9\tan \theta - 3{{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}{\text{ }}\left[ {\because \tan \dfrac{\pi }{4} = 1} \right] \\
   \Rightarrow \dfrac{{1 + \tan \theta }}{{1 - \tan \theta }} = \dfrac{{9\tan \theta - 3{{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }} \\
   \Rightarrow \left( {1 + \tan \theta } \right)\left( {1 - 3{{\tan }^2}\theta } \right) = \left( {9\tan \theta - 3{{\tan }^3}\theta } \right)\left( {1 - \tan \theta } \right) \\
   \Rightarrow 1 - 3{\tan ^2}\theta + \tan \theta - 3{\tan ^3}\theta = 9\tan \theta - 3{\tan ^3}\theta - 9{\tan ^2}\theta - 3{\tan ^4}\theta \\
   \Rightarrow 3{\tan ^4}\theta + 6{\tan ^2}\theta - 8\tan \theta + 1 = 0 \\
\]
Since, \[\alpha ,\beta ,\gamma ,\delta \] are the roots of the equation \[3{\tan ^4}\theta + 6{\tan ^2}\theta - 8\tan \theta + 1 = 0\], we have
Sum of the roots \[ = \tan \alpha + \tan \beta + \tan \gamma + \tan \delta = - \dfrac{{{\text{coefficient of ta}}{{\text{n}}^3}\theta }}{{{\text{coefficient of ta}}{{\text{n}}^4}\theta }} = - \dfrac{0}{3} = 0\]
Therefore, the value of \[\tan \alpha + \tan \beta + \tan \gamma + \tan \delta = 0\]
Thus, the correct option is D. 0

Note: For a bi-quadratic equation \[a{x^4} + b{x^3} + c{x^2} + dx + e = 0\] the sum of the roots is given by \[ - \dfrac{b}{a}\]. Here we have used the trigonometric ratio formula \[\tan \left( {a + b} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\] and \[\tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}\].