# Inverse Trigonometry – Theorem

## 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$$\frac{x+y}{1-xy}$$.