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Answer 1

Answer:

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Step-by-step explanation:

$to \: find = \\ value \: of \: x \\ \\ so \: here \\ \\ tan {}^{ - 1} \: ( \frac{x - 2}{x - 1} ) + tan {}^{ - 1} \: ( \frac{x + 2}{x + 1} ) = \frac{\pi}{4} \\ \\ tan {}^{ - 1} ( \frac{x - 2}{x - 1} \: + \: \frac{x + 2}{x + 1} \: ) = 45 \\ \\ \frac{x - 2}{x - 1} + \frac{x + 2}{x + 1} = \frac{45}{tan {}^{ - 1} } \\ \\ \frac{(x - 2)(x + 1) + (x + 2)(x - 1)}{(x + 1)(x - 1)} = tan \: 45 \\ \\ \frac{x {}^{2} + x - 2x - 2 + x {}^{2} - x + 2x - 2 }{x {}^{2} - 1} = 1 \\ \\ 2x {}^{2} - x + x - 4 = x {}^{2} - 1 \\ \\ 2x {}^{2} - x {}^{2} = - 1 + 4 \\ \\ x {}^{2} = 3 \\ \\ taking \: squre \: roots \: on \: both \: sides \\ we \: get \\ \\ x = ± \: \sqrt{3}$