Question

pls solve it ....,,,,,​

Answer 1

${tan}^{ - 1} ( \frac{x - 2}{x - 1} ) + {tan}^{ - 1} ( \frac{x + 2}{x + 1} ) = \frac{\pi}{4} \\ \implies {tan}^{ - 1} ( \frac{ \frac{x - 2}{x - 1} + \frac{x + 2}{x + 1} }{1 - \frac{x - 2}{x - 1} . \frac{x + 2}{x + 1} } ) = \frac{\pi}{4} \\ \implies ( \frac{(x - 2)(x +1 ) + (x + 2)(x - 1)}{ ({x}^{2} - 1) - ( {x}^{2} - 4)} ) = tan\frac{\pi}{4} \\ \implies \frac{ {x}^{2} - x - 2 + {x}^{2} + x - 2 }{3} = 1 \\ \implies \: 2 {x}^{2} - 4 = 3 \\ \implies \: 2{x}^{2} = 7 \\ \implies \: {x}^{2} = \frac{7}{2} \\ \implies \: x = ±\sqrt{ \frac{7}{2} }$