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We will learn how to prove the property of the inverse trigonometric function 3 arctan(x) = arctan(3x−x31−3x2) or, 3 tan−1 x = tan−1 (3x−x31−3x2).
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
11 and 12 Grade Math
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