Question

$\tan \mathrm{\theta }\tan \left(\mathrm{\theta }+60°\right)+\tan \mathrm{\theta }\tan \left(\mathrm{\theta }-60°\right)+\tan \left(\mathrm{\theta }+60°\right)\tan \left(\mathrm{\theta }-60°\right)=-3$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\mathrm{tan\theta }\tan \left(\mathrm{\theta }+60°\right)+\mathrm{tan\theta }\tan \left(\mathrm{\theta }-60°\right)+\tan \left(\mathrm{\theta }+60°\right)\tan \left(\mathrm{\theta }-60°\right) \\ =\mathrm{tan\theta }\left(\frac{\mathrm{tan\theta }+\sqrt{3}}{1-\sqrt{3}\mathrm{tan\theta }}\right)+\mathrm{tan\theta }\left(\frac{\mathrm{tan\theta }-\sqrt{3}}{1+\sqrt{3}\mathrm{tan\theta }}\right)+\left(\frac{\mathrm{tan\theta }+\sqrt{3}}{1-\sqrt{3}\mathrm{tan\theta }}\right)\left(\frac{\mathrm{tan\theta }-\sqrt{3}}{1+\sqrt{3}\mathrm{tan\theta }}\right) \\ \left[\tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}\mathrm{and}\tan \left(A-B\right)=\frac{\tan A-\tan B}{1+\tan A\tan B}\right] \\ =\frac{\left(1+\sqrt{3}\mathrm{tan\theta }\right)\mathrm{tan\theta }\left(\mathrm{tan\theta }+\sqrt{3}\right)+\left(1-\sqrt{3}\mathrm{tan\theta }\right)\mathrm{tan\theta }\left(\mathrm{tan\theta }-\sqrt{3}\right)+{\tan }^2\mathrm{\theta }-3}{\left(1+\sqrt{3}\mathrm{tan\theta }\right)\left(1-\sqrt{3}\mathrm{tan\theta }\right)} \\ =\frac{\left(\mathrm{tan\theta }+\sqrt{3}{\tan }^2\mathrm{\theta }\right)\left(\mathrm{tan\theta }+\sqrt{3}\right)+\left(\mathrm{tan\theta }-\sqrt{3}\mathrm{tan\theta }\right)\left(\mathrm{tan\theta }-\sqrt{3}\right)+{\tan }^2\mathrm{\theta }-3}{1-3{\tan }^2\mathrm{\theta }} \end{gathered}$$

$$\begin{gathered} =\frac{\left({\tan }^2\mathrm{\theta }+\sqrt{3}{\tan }^3\mathrm{\theta }+\sqrt{3}\mathrm{tan\theta }+3{\tan }^2\mathrm{\theta }\right)+\left({\tan }^2\mathrm{\theta }-\sqrt{3}{\tan }^3\mathrm{\theta }-\sqrt{3}\mathrm{tan\theta }+3{\tan }^2\mathrm{\theta }\right)+{\tan }^2\mathrm{\theta }-3}{1-3{\tan }^2\mathrm{\theta }} \\ =\frac{9{\tan }^2\mathrm{\theta }-3}{1-3{\tan }^2\mathrm{\theta }} \\ =\frac{-3\left(1-3{\tan }^2\mathrm{\theta }\right)}{\left(1-3{\tan }^2\mathrm{\theta }\right)} \\ =-3=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$