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प्रश्न
Solve for x : \[\tan^{- 1} \left( \frac{x - 2}{x - 1} \right) + \tan^{- 1} \left( \frac{x + 2}{x + 1} \right) = \frac{\pi}{4}\] .
उत्तर
\[\tan^{- 1} \left( \frac{x - 2}{x - 1} \right) + \tan^{- 1} \left( \frac{x + 2}{x + 1} \right) = \frac{\pi}{4}\]
\[ \Rightarrow \tan^{- 1} \left( \frac{x - 2}{x - 1} \right) + \tan^{- 1} \left( \frac{x + 2}{x + 1} \right) = \tan^{- 1} 1\]
\[ \Rightarrow \tan^{- 1} \left( \frac{x - 2}{x - 1} \right) = \tan^{- 1} 1 - \tan^{- 1} \left( \frac{x + 2}{x + 1} \right)\]
\[ \Rightarrow \tan^{- 1} \left( \frac{x - 2}{x - 1} \right) = \tan^{- 1} \left( \frac{1 - \frac{x + 2}{x + 1}}{1 + \frac{x + 2}{x + 1}} \right)\]
\[ \Rightarrow \tan^{- 1} \left( \frac{x - 2}{x - 1} \right) = \tan^{- 1} \left( \frac{x + 1 - x - 2}{x + 1 + x + 2} \right)\]
\[ \Rightarrow \tan^{- 1} \left( \frac{x - 2}{x - 1} \right) = \tan^{- 1} \left( \frac{- 1}{2x + 3} \right)\]
\[ \Rightarrow \frac{x - 2}{x - 1} = \frac{- 1}{2x + 3}\]
\[ \Rightarrow 2 x^2 + 3x - 4x - 6 = - x + 1\]
\[ \Rightarrow 2 x^2 = 1 + 6\]
\[ \Rightarrow x^2 = 7\]
\[ \Rightarrow x = \pm \sqrt{\frac{7}{2}}\]
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