Question

Prove the following:

3tan⁶10° – 27 tan⁴10° + 33tan²10° = 1

Answer 1

Since, tan30° = `1/sqrt(3)`

∴ tan3(10°) = `1/sqrt(3)`

∴ `(3tan10^circ - tan^3 10^circ)/(1 - 3tan^2 10^circ) = 1/sqrt(3)`

Squaring both the sides, we get

`((3tan10^circ - tan^3 10^circ)^2)/((1 - 3tan^2  10^circ)^2) = 1/3`

∴ `(9tan^2 10^circ - 6tan^4 10^circ + tan^6 10^circ)/(1 - 6tan^2 10^circ + 9tan^4 10^circ) = 1/3`

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

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

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