Question

with explanation and proper steps please solve this question

Answer 1

substitutin thetha 20

tan(3x) = (3 tan(x) − tan³(x)) / (1 − 3 tan²(x))

(3 tan(π/9) − tan³(π/9)) / (1 − 3 tan²(π/9)) = tan(π/3)

(3 tan(π/9) − tan³(π/9)) / (1 − 3 tan²(π/9)) = √3

(3 tan(π/9) − tan³(π/9)) = √3 (1 − 3 tan²(π/9))

Square both sides:

(3 tan(π/9) − tan³(π/9))² = 3 (1 − 3 tan²(π/9))²

tan⁶(π/9) − 6 tan⁴(π/9) + 9 tan²(π/9) = 3 (9 tan⁴(π/9) − 6 tan²(π/9) + 1)

tan⁶(π/9) − 6 tan⁴(π/9) + 9 tan²(π/9) = 27 tan⁴(π/9) − 18 tan²(π/9) + 3

tan⁶(π/9) − 6 tan⁴(π/9) − 27 tan⁴(π/9) + 9 tan²(π/9) + 18 tan²(π/9) = 3

tan⁶(π/9) − 33 tan⁴(π/9) + 27 tan²(π/9) = 3