Question

Prove that:

tan^–1x + tan^–1y = `π + tan^-1((x + y)/(1 - xy))`, provided x > 0, y > 0, xy > 1

Answer 1

Let tan^–1x = α and tan^–1y = β

∴ tan α = x where – π/2 < α π/2

and tan β = y where – π/2 < β < π/2

Since x > 0, y > 0, therefore 0 < α < π/2 and 0 < β < π/2

Since x > `0, y > 0 and xy > 1, (x + y)/(1 - xy) < 0`

∴ tan (α + β) = `(tan α + tanβ)/(1 - tan α . tanβ)`

∴ tan (α + β) = `(x + y)/(1 - xy) < 0`

Now, 0 < α < π/2, 0 < β < π/2

∴ 0 < α + β < π

But tan (α + β) < 0

∴ π/2 < (α + β) < π     ...[∵ tan θ < 0 in the second quadrant]

But inverse of tangent does not exist, if π/2 < (α + β) < π

∴ π/2 – π < (α + β) – π < π – π

∴ – π/2 < (α + β) – π < 0

Consider, tan (α + β – π) = – tan [π – (α + β)] = tan (α + β)

∴ tan (α + β – π) = `(x + y)/(1 - xy)`

∴ `α + β - π = tan^-1((x + y)/(1 - xy))`

∴ `α + β = π + tan^-1((x + y)/(1 - xy))`

∴ tan^–1x + tan^–1y = `π + tan^-1((x + y)/(1 - xy))`.