3 arctan(x) = arctan(\(\frac{3x - x^{3}}{1 - 3 x^{2}}\))

We will learn how to prove the property of the inverse trigonometric function 3 arctan(x) = arctan(\(\frac{3x - x^{3}}{1 - 3 x^{2}}\)) or, 3 tan\(^{-1}\) x = tan\(^{-1}\) (\(\frac{3x - x^{3}}{1 - 3x^{2}}\)).

Proof:  

Let, tan\(^{-1}\) x = θ      

Therefore, tan θ = x

Now we know that, tan 3θ = \(\frac{3 tan θ - tan^{3}θ}{1 - 3 tan^{2}θ}\)

⇒ tan 3θ = \(\frac{3x - x^{3}}{1 - 3x^{2}}\)

Therefore, 3θ = tan\(^{-1}\) (\(\frac{3x - x^{3}}{1 - 3x^{2}}\))

⇒ 3 tan\(^{-1}\) x = tan\(^{-1}\) (\(\frac{3x - x^{3}}{1 - 3x^{2}}\))

or, 3 arctan(x) = arctan(\(\frac{3x - x^{3}}{1 - 3x^{2}}\)).           Proved.




















11 and 12 Grade Math

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