Question

Solve for X

tan^-1(2x)+tan^-1 (3x) = π/4 ​

Answer 1

Hii

Step-by-step explanation:

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Solution is in this attachment

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Answer 2

Step-by-step explanation:

$we \: know \: that \\ tan {}^{ - 1} x + tan {}^{ - 1} y = tan {}^{ - 1} ( \frac{x + y}{1 - xy} ) \\ replacing \: x \: by \: 2x \: and \: y \: by \: 3x. \\ tan {}^{ - 1} 2x + tan {}^{ - 1} 3y = tan {}^{ - 1} ( \frac{2x +3 x}{1 - 2x \times 3x} ) \\ tan { }^{ - 1} = ( \frac{5x}{1 - 6x {}^{2} } ) \\ now \: given \\ tan {}^{ - 1} 2x + tan {}^{ - 1}3 y = \frac{\pi}{4} \\ tan {}^{ - 1} ( \frac{5x}{ 1 - 6x {}^{2} } ) = \frac{\pi}{4} \\ \frac{5x}{1 - 6x {}^{2} } = tan \frac{\pi}{4} \\ \\ \frac{5x}{1 - 6x^2} = 1. \\ 5x = 1 \times (1 - 6x^2) \\ 5x = 1 - 6x {}^{2} \\6x {}^{2} + 5x - 1 = 0 \\ 6x { }^{2} + 6x - x - 1 = 0 \\ 6x(x + 1) - 1(x - 1) = 0 \\ (6x - 1)(x + 1) = 0 \\ 6x - 1 = 0 \: \: and \: \: x + 1 = 0 \\ x = \frac{1}{6} \: \: and \: \: x = - 1$

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