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

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

• $\frac{\pi}{2}$

• $\frac{\pi}{2}$

• $\frac{\pi}{2}$

• $\frac{\pi}{2}$

Solution

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

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

Hence, the correct answer is option D.

Concept: Transformation Formulae
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 8 Transformation formulae
Q 14 | Page 22