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Question
It \[\tan^{- 1} \frac{x + 1}{x - 1} + \tan^{- 1} \frac{x - 1}{x} = \tan^{- 1}\] (−7), then the value of x is
Options
0
−2
1
2
Solution
(d) 2
We know that
\[\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} \left( \frac{x + y}{1 - xy} \right)\]
\[\therefore \tan^{- 1} \left( \frac{x + 1}{x - 1} \right) + \tan^{- 1} \left( \frac{x - 1}{x} \right) = \tan^{- 1} \left( - 7 \right)\]
\[ \Rightarrow \tan^{- 1} \left( \frac{\frac{x + 1}{x - 1} + \frac{x - 1}{x}}{1 - \frac{x + 1}{x - 1} \times \frac{x - 1}{x}} \right) = \tan^{- 1} \left( - 7 \right)\]
\[ \Rightarrow \tan^{- 1} \left( \frac{\frac{x^2 + x + x^2 - 2x + 1}{x\left( x - 1 \right)}}{\frac{x^2 - x - x^2 + 1}{x\left( x - 1 \right)}} \right) = \tan^{- 1} \left( - 7 \right)\]
\[ \Rightarrow \tan^{- 1} \left( \frac{2 x^2 - x + 1}{- x + 1} \right) = \tan^{- 1} \left( - 7 \right)\]
So, we get
\[\frac{2 x^2 - x + 1}{- x + 1} = - 7\]
\[ \Rightarrow 2 x^2 - x + 1 = 7x - 7\]
\[ \Rightarrow 2 x^2 - 8x + 8 = 0\]
\[ \Rightarrow x^2 - 4x + 4 = 0\]
\[ \Rightarrow \left( x - 2 \right)^2 = 0\]
\[ \Rightarrow x = 2\]
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