Solve for x- 2tan-12x1-x2-

Given, 2tan^ 12x1 x^2=π 2× 2tan^ 1 x=π [ Using direct formula for 2tan^ 1x ] 4tan^ 1x=π tan^ 1 x=π4 x=tanπ4 x=1 . ...

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Domain and Range of Basic Inverse Trigonometric Functions

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Question

Solve for $x$ :
$2 {tan}^{- 1} \frac{2 x}{1 - x^{2}} = π$ .

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Solution

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3 {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) - 4 {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) + 2 {tan}^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{π}{3}

3 s i n^{- 1} \frac{2 x}{1 + x^{2}} - 4 c o s^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 t a n^{- 1} \frac{2 x}{1 + x^{2}} = \frac{π}{3}

x : {cos}^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}) + \frac{1}{2} {tan}^{- 1} \frac{2 x}{1 - x^{2}} = \frac{2 π}{3}

3 {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) - 4 {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) + 2 {tan}^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{π}{3}

x

3 s i n^{- 1} (\frac{2 x}{1 + x^{2}}) - 4 c o s^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) + 2 t a n^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{π}{3}

| x | < 1

x

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