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.*

**●** Inverse Trigonometric Functions

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