Question

If ${\theta }_1,{\theta }_2,{\theta }_3,{\theta }_4$ be the four values of $\theta$ which satisfy the equation $3\tan 3\theta =\tan ({45}^o+\theta )$ then prove that $\sum \tan {\theta }_1=0$

Answer 1

If $\tan \theta =t$, then the 4th degree equation in $t$ is $3t^4-6t^2+8t-1=0$
$\therefore \sum t_1=0$. The term of $t^3$ is missing