Let tan - 1 y = tan - 1 x + tan - 1 2 x 1 - x 2 , where x < 1 3 ,Then a value of y is

Given tan - 1 y = tan - 1 x + tan - 1 2 x 1 - x 2 , x < 1 3 = tan - 1 x + 2 tan - 1 x = 3 tan - 1 x ⇒ tan - 1 y = tan - 1 3 x - x 3 1 - 3 x 2 ∴ y = 3 x - x 3 1 - 3 x 2

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Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

MathematicsInverse Trigonometric FunctionsT inverse T propertyMedium2 minPYQ_2015

MathematicsMediumsingle choice

Let $\tan^{- 1} y = \tan^{- 1} x + \tan^{- 1} (\frac{2 x}{1 - x^{2}})$ , where $|x| < \frac{1}{\sqrt{3}}$ ,Then a value of $y$ is

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Solution:

Given $\tan^{- 1} y = \tan^{- 1} x + \tan^{- 1} (\frac{2 x}{1 - x^{2}}), |x| < \frac{1}{\sqrt{3}}$

$= \tan^{- 1} x + 2 \tan^{- 1} x$

$= 3 \tan^{- 1} x$

$\Rightarrow \tan^{- 1} y = \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}})$

$∴ y = \frac{3 x - x^{3}}{1 - 3 x^{2}}$

Stream:JEESubject:MathematicsTopic:Inverse Trigonometric FunctionsSubtopic:T inverse T property

⏱ 2mℹ️ Source: PYQ_2015

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