# arctan(x) + arctan(y) + arctan(z) = arctan$$\frac{x + y + z - xyz}{1 - xy - yz - zx}$$

We will learn how to prove the property of the inverse trigonometric function arctan(x) + arctan(y) + arctan(z) = arctan$$\frac{x + y + z - xyz}{1 - xy - yz - zx}$$ (i.e., tan$$^{-1}$$ x + tan$$^{-1}$$ y + tan$$^{-1}$$ z = tan$$^{-1}$$ $$\frac{x + y + z - xyz}{1 - xy - yz - zx}$$)

Prove that, tan$$^{-1}$$ x + tan$$^{-1}$$ y + tan$$^{-1}$$ z = tan$$^{-1}$$ $$\frac{x + y + z – xyz}{1 – xy – yz – zx}$$

Proof :

Let, tan$$^{-1}$$ x = α, tan$$^{-1}$$ y = β and tan$$^{-1}$$γ

Therefore, tan α = x, tan β = y and tan γ = z

We know that, tan (α + β + γ) = $$\frac{tan α + tan β + tan γ - tan α tan β tan γ}{1 - tan α tan β - tan β tan γ - tan γ tan α}$$

tan (α + β + γ)   = $$\frac{x + y + z – xyz}{1 – xy – yz – zx}$$

α + β + γ = tan$$^{-1}$$ $$\frac{x + y + z – xyz}{1 – xy – yz – zx}$$

or, tan$$^{-1}$$ x + tan$$^{-1}$$ y + tan$$^{-1}$$ z = tan$$^{-1}$$ $$\frac{x + y + z – xyz}{1 – xy – yz – zx}$$.                Proved.

Second method:

We can prove tan$$^{-1}$$ x + tan$$^{-1}$$ y + tan$$^{-1}$$ z = tan$$^{-1}$$ $$\frac{x + y + z – xyz}{1 – xy – yz – zx}$$ in other way.

We know that, tan$$^{-1}$$ x + tan$$^{-1}$$ y = tan$$^{-1}$$ $$\frac{x + y}{1 – xy}$$

Therefore, tan$$^{-1}$$ x + tan$$^{-1}$$ y + tan$$^{-1}$$ z = tan$$^{-1}$$ $$\frac{x + y}{1 – xy}$$ + tan$$^{-1}$$ z

tan$$^{-1}$$ x + tan$$^{-1}$$ y + tan$$^{-1}$$ z = tan$$^{-1}$$ $$\frac{\frac{x + y}{1 – xy} + z}{1 - \frac{x + y}{1 - xy } ∙ z}$$

tan$$^{-1}$$ x + tan$$^{-1}$$ y + tan$$^{-1}$$ z = tan$$^{-1}$$ $$\frac{x + y + z – xyz}{1 – xy – yz – zx}$$.          Proved.

Inverse Trigonometric Functions

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