Inverse Trigonometry – Theorem
Statement: Tan⁻¹ x + Tan⁻¹ y = Tan⁻¹ \(\left( \frac{x+y}{1-xy} \right)\) for x > 0, y > 0, xy < 1 = π + Tan⁻¹ \(\left( \frac{x+y}{1-xy} \right)\) for x > 0, y > 0, xy > 1.
Proof:
Case (i): Suppose x > 0, y > 0, xy < 1
x > 0 → 0 < Tan⁻¹ x < π/2
y > 0 → 0 < Tan⁻¹ y < π/2
Let Tan⁻¹ x = α, Tan⁻¹ y = β
Then x = Tan α, y = Tan β
0 < Tan⁻¹ x < π/2 → 0 < α < π/2
0 < Tan⁻¹ y < π/2 → 0 < β < π/2
0 < α < π/2, 0 < β < π/2 → 0 < α + β < π
\(\tan \left( \alpha +\beta \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\frac{\operatorname{Tan}\alpha +\operatorname{Tan}\beta }{1-\operatorname{Tan}\alpha \operatorname{Tan}\beta }=\frac{x+y}{1-xy}>0\).
0 < α + β < π, tan (α + β) > 0
0 < α + β < π / 2
∴ Tan (α + β) = tan (α + β)
= \(\frac{x+y}{1-xy}\Rightarrow {{\operatorname{Tan}}^{-1}}\frac{x+y}{1-xy}\).
= α + β = Tan⁻¹ x + Tan⁻¹ y
Case (ii): Suppose x > 0, y > 0, xy > 1
x > 0 → 0 < Tan⁻¹ x < π/2
y > 0 → 0 < Tan⁻¹ y < π/2
Let Tan⁻¹ x = α, Tan⁻¹ y = β.
Then x = Tan α, y = Tan β
0 < Tan⁻¹ x < π / 2, 0 < Tan⁻¹ y < π/2
0 < α < π/2, 0 < β < π/2
0 < α + β < π
\(\tan \left( \alpha +\beta \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\frac{\operatorname{Tan}\alpha +\operatorname{Tan}\beta }{1-\operatorname{Tan}\alpha \operatorname{Tan}\beta }=\frac{x+y}{1-xy}<0\).
0 < α + β < π, tan (α + β – π)
= -tan [π – (α + β)]
= tan (α + β) \(=\frac{x+y}{1-xy}\).
Tan⁻¹\(\frac{x+y}{1-xy}\) = α + β – π
α + β = π + Tan⁻¹\(\frac{x+y}{1-xy}\).
Tan⁻¹ x + Tan⁻¹ y = π + Tan⁻¹\(\frac{x+y}{1-xy}\).