Question

solve tan^-1(2x)+tan^-1(3x)=pie\4​

Answer 1

hope this will be helpful

Answer 2

Solution

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$\bf \large \: we \: have \: { \tan }^{ - 1} 2x + { \tan}^{ - 1} 3x = \frac{\pi}{4} \\ \: \\ \bf \: { \tan }^{ - 1} ( \frac{2x + 3x}{1 - 2x \times 3x} ) = \frac{\pi}{4} \\ \\ \bf { \tan}^{ - 1} ( \frac{5x}{1 - {6x}^{2} } = \frac{\pi}{4} \\ \\ \bf \frac{5x}{?1 - {6x}^{2} } = \tan( \frac{\pi}{4} ) = 1 \\ \\ \bf {6x}^{2} + 5x - 1 = 0 \\ \bf x = \frac{1}{6} or \: x = - 1$

since X =-1 does not satisfy the equation

as lhs of the equation become negative.

x=1/6 is the only solution of given equation