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If Tan α = X X + 1 and Tan β = 1 2 X + 1 , Then Tan β = 1 2 X + 1 is Equal to - Mathematics

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प्रश्न

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

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

 

विकल्प

  • \[\frac{\pi}{2}\]

     

  • \[\frac{\pi}{2}\]

     

  • \[\frac{\pi}{2}\]

     

  • \[\frac{\pi}{2}\]

     

MCQ
योग
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उत्तर

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

Now,
\[\tan\left( \alpha + \beta \right) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha \tan\beta}\]
\[ = \frac{\frac{x}{x + 1} + \frac{1}{2x + 1}}{1 - \frac{x}{x + 1} \times \frac{1}{2x + 1}}\]
\[ = \frac{\frac{x\left( 2x + 1 \right) + x + 1}{\left( x + 1 \right)\left( 2x + 1 \right)}}{\frac{\left( x + 1 \right)\left( 2x + 1 \right) - x}{\left( x + 1 \right)\left( 2x + 1 \right)}}\]
\[ = \frac{2 x^2 + x + x + 1}{2 x^2 + 3x + 1 - x}\]
\[= \frac{2 x^2 + 2x + 1}{2 x^2 + 2x + 1}\]
\[ = 1\]
\[\therefore \tan\left( \alpha + \beta \right) = 1 = \tan\frac{\pi}{4}\]
\[ \Rightarrow \alpha + \beta = \frac{\pi}{4}\]

Hence, the correct answer is option D.

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Transformation Formulae
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Transformation formulae - Exercise 8.4 [पृष्ठ २२]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 8 Transformation formulae
Exercise 8.4 | Q 14 | पृष्ठ २२

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