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Solve for X : Tan − 1 ( X − 2 X − 1 ) + Tan − 1 ( X + 2 X + 1 ) = π 4 . - Mathematics

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Question

Solve for x : \[\tan^{- 1} \left( \frac{x - 2}{x - 1} \right) + \tan^{- 1} \left( \frac{x + 2}{x + 1} \right) = \frac{\pi}{4}\] .

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Solution

\[\tan^{- 1} \left( \frac{x - 2}{x - 1} \right) + \tan^{- 1} \left( \frac{x + 2}{x + 1} \right) = \frac{\pi}{4}\]

\[ \Rightarrow \tan^{- 1} \left( \frac{x - 2}{x - 1} \right) + \tan^{- 1} \left( \frac{x + 2}{x + 1} \right) = \tan^{- 1} 1\]

\[ \Rightarrow \tan^{- 1} \left( \frac{x - 2}{x - 1} \right) = \tan^{- 1} 1 - \tan^{- 1} \left( \frac{x + 2}{x + 1} \right)\]

\[ \Rightarrow \tan^{- 1} \left( \frac{x - 2}{x - 1} \right) = \tan^{- 1} \left( \frac{1 - \frac{x + 2}{x + 1}}{1 + \frac{x + 2}{x + 1}} \right)\]

\[ \Rightarrow \tan^{- 1} \left( \frac{x - 2}{x - 1} \right) = \tan^{- 1} \left( \frac{x + 1 - x - 2}{x + 1 + x + 2} \right)\]

\[ \Rightarrow \tan^{- 1} \left( \frac{x - 2}{x - 1} \right) = \tan^{- 1} \left( \frac{- 1}{2x + 3} \right)\]

\[ \Rightarrow \frac{x - 2}{x - 1} = \frac{- 1}{2x + 3}\]

\[ \Rightarrow 2 x^2 + 3x - 4x - 6 = - x + 1\]

\[ \Rightarrow 2 x^2 = 1 + 6\]

\[ \Rightarrow x^2 = 7\]

\[ \Rightarrow x = \pm \sqrt{\frac{7}{2}}\]

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2015-2016 (March) Foreign Set 2

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