Question

Prove the following:

3tan⁶10° – 27 tan⁴10° + 33tan²10° = 1

Answer 1

Since, tan30° = ${\displaystyle \frac{1}{\sqrt{3}}}$

∴ tan3(10°) = ${\displaystyle \frac{1}{\sqrt{3}}}$

∴ ${\displaystyle \frac{3{\tan 10}^{\circ }-{\tan }^3{10}^{\circ }}{1-3{\tan }^2{10}^{\circ }}=\frac{1}{\sqrt{3}}}$

Squaring both the sides, we get

${\displaystyle \frac{{\left(3{\tan 10}^{\circ }-{\tan }^3{10}^{\circ }\right)}^2}{{(1-3{\tan }^2 {10}^{\circ })}^2}=\frac{1}{3}}$

∴ ${\displaystyle \frac{9{\tan }^2{10}^{\circ }-6{\tan }^4{10}^{\circ }+{\tan }^6{10}^{\circ }}{1-6{\tan }^2{10}^{\circ }+9{\tan }^4{10}^{\circ }}=\frac{1}{3}}$

∴ 3(9tan²10° – 6tan⁴10°+ tan⁶10°) = 1 – 6tan²10° + 9tan⁴10°

∴ 27tan²10° – 18tan⁴10° + 3tan⁶10° = 1 – 6tan²10° + 9tan⁴10°

∴ 3tan⁶10° – 27tan⁴10° + 33tan²10° = 1