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
11 and 12 Grade Math
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