Question
Prove \tan20\degree -\tan40\degree +\tan80\degree =3\sqrt{3}
Prove \tan20\degree -\tan40\degree +\tan80\degree =3\sqrt{3}
Given
\tan20\degree -\tan40\degree +\tan80\degree =3\sqrt{3}
Let's convert the LHS in terms of one angle 20°.
\tan20\degree -\tan(60°-20\degree) +\tan(60\degree+20°)
Using the sum identities for tangent function, the expression is converted to
\tan(\alpha+\beta) = \dfrac{\tan \alpha + \tan \beta }{1-\tan \alpha\tan \beta}
\tan(\alpha -\beta) = \dfrac{\tan \alpha - \tan \beta }{1+\tan \alpha\tan \beta}
\tan20\degree - \dfrac{\tan 60° - \tan 20° }{1+\tan 60°\tan 20°} + \dfrac{\tan 60° + \tan 20° }{1-\tan 60°\tan 20°}
Let t = \tan20°, then the expression is simplified as
t- \dfrac{\sqrt{3}-t }{1+\sqrt{3}t } + \dfrac{\sqrt{3}+t }{1-\sqrt{3}t }
=t - \dfrac{(\sqrt{3}-t)(1-\sqrt{3}t)-(\sqrt{3}+t)(1+\sqrt{3}t )}{(1+\sqrt{3}t)(1-\sqrt{3}t)}
=t-\dfrac{\sqrt{3}-3t-t+\sqrt{3}t^2-(\sqrt{3} + 3t+t+\sqrt{3}t^2 ) }{1-3t^2}
=t-\dfrac{-8t }{1-3t^2}
=\dfrac{t-3t^3+8t}{1-3t^2}
=\dfrac{9t-3t^3}{1-3t^2}
=3\cdot \dfrac{3t-t^3}{1-3t^2}
Substitute t back to \tan20°, and then use triple identity for tangent function
=3\cdot \dfrac{3 \tan20°- \tan^320°}{1-3 \tan^220°}
=3\cdot \tan60\degree = 3\sqrt{3}