Question

Let $\tan \,\alpha = \frac{a}{a+1}$ and $\tan \,\beta = \frac{1}{2a + 1}$ then $\alpha + \beta$ is

• A. $\frac{\pi}{4}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{2}$
• D. $\pi$

Answer 1

Answer: (A)

$\frac{\pi}{4}$

Answer 2

Given, $\tan \alpha =\frac{a}{a+1}$ and $\tan \beta=\frac{1}{2 a+1}$
$\therefore \tan (\alpha+\beta) =\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}$
$=\frac{\frac{a}{a+1}+\frac{1}{2 a+1}}{1-\frac{a}{(a+1)} \times \frac{1}{(2 a+1)}}$
$=\frac{a(2 a+1)+(a+1)}{(a+1)(2 a+1)-a}$
$=\frac{2 a^{2}+2 a+1}{2 a^{2}+2 a+1}=1$
$\Rightarrow \alpha+\beta=\frac{\pi}{4}$