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.






11 and 12 Grade Math

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