Let tan alpha = frac{a}{a+1} and tan beta = f | vedclass

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Question

(A) We use the formula $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$.
Substituting the given values:
$	an(\a

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(A) We use the formula $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$.
Substituting the given values:
$	an(\alpha + \beta) = \frac{\frac{a}{a+1} + \frac{1}{2a+1}}{1 - \left(\frac{a}{a+1}ight) \left(\frac{1}{2a+1}ight)}$
$= \frac{\frac{a(2a+1) + 1(a+1)}{(a+1)(2a+1)}}{\frac{(a+1)(2a+1) - a}{(a+1)(2a+1)}}$
$= \frac{2a^2 + a + a + 1}{2a^2 + a + 2a + 1 - a}$
$= \frac{2a^2 + 2a + 1}{2a^2 + 2a + 1} = 1$
Since $	an(\alpha + \beta) = 1$, we have $\alpha + \beta = \frac{\pi}{4}$.

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

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