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Question
The number of solutions of the equation \[\tan^{- 1} 2x + \tan^{- 1} 3x = \frac{\pi}{4}\] is
Options
2
3
1
none of these
Solution
(a) 2
We know that
\[\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} \left( \frac{x + y}{1 - xy} \right)\]
\[\therefore \tan^{- 1} 2x + \tan^{- 1} 3x = \frac{\pi}{4}\]
\[ \Rightarrow \tan^{- 1} \left( \frac{2x + 3x}{1 - 2x \times 3x} \right) = \frac{\pi}{4}\]
\[ \Rightarrow \frac{2x + 3x}{1 - 2x \times 3x} = \tan\frac{\pi}{4}\]
\[ \Rightarrow \frac{5x}{1 - 6 x^2} = 1 \]
\[ \Rightarrow 5x = 1 - 6 x^2 \]
\[ \Rightarrow 6 x^2 + 5x - 1 = 0\]
Therefore, there are two solutions.
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