Prove \tan20\degree -\tan40\degree +\tan80\degree =3\sqrt{3}

Collected in the board: Trigonometry

Steven Zheng posted 7 hours ago



\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}



=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}

Steven Zheng posted 7 hours ago

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