Question

Let $\alpha ,\beta$ are acute and $\tan \alpha =\frac{a}{a+1}$, $\tan \beta =\frac{1}{2a+1}$, then find the value of $\alpha +\beta$.

Answer 1

Given, $\alpha ,\beta$ are acute and $\tan \alpha =\frac{a}{a+1}$......(1), $\tan \beta =\frac{1}{2a+1}$......(2).

Now,

$\tan (\alpha +\beta )$

$=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha .\tan \beta }$

$=\frac{\frac{a}{a+1}+\frac{1}{2a+1}}{1-\frac{a}{a+1}.\frac{1}{2a+1}}$ [ Using (1) and (2)]

$=\frac{2a^2+2a+1}{2a^2+3a+1-a}$

$=\frac{2a^2+2a+1}{2a^2+2a+1}$

$=1$.

or, $\tan (\alpha +\beta )=1$

or, $\alpha +\beta =\frac{\pi }{4}$.