Question

Prove that tan⁻¹1 + tan⁻¹2 + tan⁻¹3 = π.

Answer 1

By using the result

`tan^-1 x + tan^-1 y = pi + tan^-1 ((x + y)/(1 - xy))`, if x > 0, y > 0, xy > 1, we get

`tan^-1 + tan^-1 2 = pi + tan^-1 ((1 + 2)/(1 - 1 xx 2))`  ...[∵ 1 × 2 = 2 > 1]

= `pi + tan^-1 (3/(-1)) = pi + tan^-1 (-3)`

= π − tan⁻¹3  ...[∵ tan⁻¹(−θ) = −tan⁻¹θ]

∴ LHS = tan⁻¹1 + tan⁻¹2 + tan⁻¹3

= π − tan⁻¹3 + tan⁻¹3

= π = RHS