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If Tan α = X X + 1 and Tan α = X X + 1 , Then α + β is Equal to - CBSE (Science) Class 11 - Mathematics

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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

  • \[\frac{\pi}{2}\]

     

  • \[\frac{\pi}{3}\]

     

  • \[\frac{\pi}{6}\]

     

  • \[\frac{\pi}{4}\]

     

Solution

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

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

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

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

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

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

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

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

Hence, the correct answer is option D.

  Is there an error in this question or solution?

APPEARS IN

 RD Sharma Solution for Mathematics Class 11 (2019 to Current)
Chapter 7: Values of Trigonometric function at sum or difference of angles
Q: 23 | Page no. 29
Solution If Tan α = X X + 1 and Tan α = X X + 1 , Then α + β is Equal to Concept: Trigonometric Functions of Sum and Difference of Two Angles.
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