Question

Tan⁶20-33Tan⁴20+27Tan²20=3​

Answer 1

Step-by-step explanation:

Given Tan⁶20-33Tan⁴20+27Tan²20=3

• If tan x = t, then tan 2x = 2t / (1 – t^2)
• Now tan 3x = tan (x + 2x) = tan x + tan 2x / 1 – tan x tan 2x
•                                        = 1 + 2t / 1 – t^2 / 1 – 2t^2 / 1 – t^2
• Now tan 3x = (3t – t^3) / (1 – 3t^2 )
• We know that tan 60 = √3, so tan 20 is true for (3t – t^3) / (1 – 3t^2) = √3
• Now squaring both sides we get
• (3t – t^3)^2 = 3(1 – 3t^2)^2
• 9t^2 – 6t^4 + t^6 = 3 (1 – 6t^2 + 9t^4)
• 9t^2 – 6t^4 + t^6 = 3 – 18t^2 + 27t^4
• 9t^2 + 18t^2 – 6t^4 – 27t^4 + t^6 = 3

t^6 – 33 t^4 + 27 t^2 = 3