Question

Formula for solving tan²

Answer 1

Answer:

tan2x= 13

⇒ tan x= ± 1√3

⇒ tan x = tan (±π6)

Therefore, x= nπ ± π6, where n = 0, ±1, ±2,…………

When, n = 0 then x = ± π6 = π6 or,- π6

If n = 1 then x = π ± π6 + 5π6 or,- 7π6

If n = -1 then x = - π ± π6 =- 7π6, - 5π6

Therefore, the required solutions in – π ≤ x ≤ π are x = π6, 5π6, - π6, - 5π6.

Answer 2

Answer:

Tan²θ = Sec²θ - 1

Tan2A = 2 ⋅ TanA / (1 - Tan²A)

TanA = 2 ⋅ Tan(A/2) / [1 - Tan²(A/2)]

Tan²(A/2) = (1 - CosA) / (1 + CosA)