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Question
If \[\tan\alpha = \frac{x}{x + 1}\] and \[\tan\alpha = \frac{x}{x + 1}\], then \[\alpha + \beta\] is equal to
Options
- \[\frac{\pi}{2}\]
- \[\frac{\pi}{3}\]
- \[\frac{\pi}{6}\]
- \[\frac{\pi}{4}\]
Solution
It is given that \[\tan\alpha = \frac{x}{x + 1}\] and
\[\tan\left( \alpha + \beta \right) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}\]
\[ = \frac{\frac{x}{x + 1} + \frac{1}{2x + 1}}{1 - \frac{x}{x + 1} \times \frac{1}{2x + 1}}\]
\[ = \frac{\frac{x\left( 2x + 1 \right) + \left( x + 1 \right)}{\left( x + 1 \right)\left( 2x + 1 \right)}}{\frac{\left( x + 1 \right)\left( 2x + 1 \right) - x}{\left( x + 1 \right)\left( 2x + 1 \right)}}\]
\[ = \frac{2 x^2 + x + x + 1}{2 x^2 + 3x + 1 - x}\]
\[= \frac{2 x^2 + 2x + 1}{2 x^2 + 2x + 1}\]
\[ = 1\]
\[\therefore \alpha + \beta = \frac{\pi}{4} \left( \tan\frac{\pi}{4} = 1 \right)\]
Hence, the correct answer is option D.
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