MCQ

Sum

If \[\tan\alpha = \frac{x}{x + 1}\] and

\[\tan\beta = \frac{1}{2x + 1}\], then

\[\tan\beta = \frac{1}{2x + 1}\] is equal to

#### Options

- \[\frac{\pi}{2}\]
- \[\frac{\pi}{2}\]
- \[\frac{\pi}{2}\]
- \[\frac{\pi}{2}\]

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#### Solution

It is given that \[\tan\alpha = \frac{x}{x + 1}\] and \[\tan\beta = \frac{x}{x + 1}\]

Now,

\[\tan\left( \alpha + \beta \right) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha \tan\beta}\]

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

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

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

\[= \frac{2 x^2 + 2x + 1}{2 x^2 + 2x + 1}\]

\[ = 1\]

\[\therefore \tan\left( \alpha + \beta \right) = 1 = \tan\frac{\pi}{4}\]

\[ \Rightarrow \alpha + \beta = \frac{\pi}{4}\]

\[\tan\left( \alpha + \beta \right) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha \tan\beta}\]

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

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

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

\[= \frac{2 x^2 + 2x + 1}{2 x^2 + 2x + 1}\]

\[ = 1\]

\[\therefore \tan\left( \alpha + \beta \right) = 1 = \tan\frac{\pi}{4}\]

\[ \Rightarrow \alpha + \beta = \frac{\pi}{4}\]

Hence, the correct answer is option D.

Concept: Transformation Formulae

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