Find the value of x which satisfy euqation : ${tan}^{- 1} 2 x + {tan}^{- 1} 3 x = π / 4$

x = - 1 / 6

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x = + 1 / 6

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x = - 1

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x = + 1

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Solution

The correct option is C $x = + 1 / 6$
Given, ${tan}^{- 1} 2 x + {tan}^{- 1} 3 x = \frac{π}{4}$
or ${tan}^{- 1} (\frac{2 x + 3 x}{1 - 6 x^{2}}) + {tan}^{- 1} = \frac{π}{4}$
$\frac{5 x}{1 - 6 x^{2}} = 1$
$6 x^{2} + 5 x - 1 = 0$ (i)
$(6 x - 1) (x + 1) = 0$
$x = 1 / 6$ or $- 1$ , but $x = - 1$ does not satisfy Eq. (i).
Hence, $x = 1 / 6$

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Find the value of x which satisfy euqation : ${tan}^{- 1} 2 x + {tan}^{- 1} 3 x = π / 4$