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

#160; tan^ 1(x 1x 2)+tan^ 1(x+1x+2)=tan^ 1(x 1x 2+x+1x+21 (x 1)(x+1)(x 2)(x+2)) #160; =tan^ 1(2x^2 4 3)=π4 #160; 2x^2 4 3=1 #160; x= ...

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Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

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Question

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

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

{t a n}^{- 1} \frac{x - 1}{x - 2} + {t a n}^{- 1} \frac{x + 1}{x + 2} = \frac{π}{4}

{t a n}^{- 1} \frac{x - 1}{x + 1} + {t a n}^{- 1} \frac{2 x - 1}{2 x + 1} = {t a n}^{- 1} \frac{23}{36}

{tan}^{- 1} \frac{x + 7}{x - 1} + {tan}^{- 1} \frac{x - 1}{x} = π - {tan}^{- 1} 7

{tan}^{- 1} (x - 1) + {tan}^{- 1} x + {tan}^{- 1} (x + 1) = {tan}^{- 1} 3 x .

\frac{3 π}{4}

\frac{8}{31}

(\frac{1 - x}{1 + x}) - \frac{1}{2}

\frac{π}{12}

(\frac{8}{79})

\tan^{- 1} \frac{x}{2} + \tan^{- 1} \frac{x}{3} = \frac{π}{4}, 0 < x < \sqrt{6}

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

\tan^{- 1} (2 + x) + \tan^{- 1} (2 - x) = \tan^{- 1} \frac{2}{3}, where x < - \sqrt{3} or, x > \sqrt{3}

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

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