Solve - tan-1x - 1x - 2 - tan-1x - 1x - 2 - -4

tan^ 1(x 1x 2)+tan^ 1(x+1x+2)=π4 tan^ 1a+tan^ 1b=tan^ 1(a+b1 ab) tan^ 1((x 1)(x 2)+(x+1)(x+2)1 (x 1x 2)(x+1x+2))=π4 ∴(x 1)(x+2)+(x+1)(x 2)(x ...

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Homogeneous Differential Equation

Solve : tan...

Question

Solve : ${tan}^{- 1} (\frac{x - 1}{x - 2}) + {tan}^{- 1} (\frac{x + 1}{x + 2}) = \frac{π}{4}$

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{tan}^{- 1} (\frac{x - 1}{x - 2}) + {tan}^{- 1} (\frac{x + 1}{x + 2}) = \frac{π}{4}

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