Question

tan⁻¹(3x-x³/1-3x²)=3tan⁻¹x,0 < x < 1/√3,prove it

Answer 1

we have to prove that,

tan⁻¹(3x-x³/1-3x²)=3tan⁻¹x, where 0 < x < 1/√3

Let 3tan^-1x = A ......(1)

⇒tan^-1x = A/3

⇒tan(A/3) = x

use formula, tan3θ= (3tanθ- tan³θ)/(1 - 3tan²θ)

so, tanA = {3tan(A/3) - tan³(A/3)}/{1 - 3tan²(A/3)}

= {3x - x³}/{1 - 3x²}

hence, tanA = (3x - x³)/(1 - 3x²)

⇒A = tan^-1[ (3x - x³)/(1 - 3x²)] .....(2)

from equations (1) and (2),

tan^-1[(3x - x³)/(1 - 3x²) ] = 3tan^-1x [proved]