Tricky Trigo

Geometry Level 3

tan ⁡ 2 0 ∘ tan ⁡ 4 0 ∘ tan ⁡ 6 0 ∘ tan ⁡ 8 0 ∘ = ? \large \tan20^\circ\tan40^\circ\tan60^\circ\tan80^\circ = \, ?


The answer is 3.

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Jus Jaisinghani by Jus Jaisinghani , 21, India

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1 solution

X = tan ⁡ 2 0 ∘ tan ⁡ 4 0 ∘ tan ⁡ 6 0 ∘ tan ⁡ 8 0 ∘ = tan ⁡ 2 0 ∘ tan ⁡ ( 6 0 ∘ − 2 0 ∘ ) tan ⁡ 6 0 ∘ tan ⁡ ( 6 0 ∘ + 2 0 ∘ ) = tan ⁡ 2 0 ∘ ( 3 − tan ⁡ 2 0 ∘ 1 + 3 tan ⁡ 2 0 ∘ ) 3 ( 3 + tan ⁡ 2 0 ∘ 1 − 3 tan ⁡ 2 0 ∘ ) = 3 tan ⁡ 2 0 ∘ ( 3 − tan ⁡ 2 2 0 ∘ ) 1 − 3 tan ⁡ 3 2 0 ∘ = 3 tan ⁡ 6 0 ∘ = 3 \begin{aligned} \mathscr X & = \tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ \\ & = \tan 20^\circ \tan(60^\circ - 20^\circ) \tan 60^\circ \tan(60^\circ + 20^\circ) \\ & = \tan 20^\circ \left(\frac {\sqrt 3 -\tan 20^\circ}{1+\sqrt 3 \tan 20^\circ} \right) \sqrt 3 \left(\frac {\sqrt 3 + \tan 20^\circ}{1-\sqrt 3 \tan 20^\circ} \right) \\ & = \frac {\sqrt 3 \tan 20^\circ (3-\tan^2 20^\circ)}{1-3\tan^3 20^\circ} \\ & = \sqrt 3 \tan 60^\circ \\ & = \boxed{3} \end{aligned}

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