If tan alpha = frac{m}{m + 1} and tan beta = | vedclass

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(B) We have $	an \alpha = \frac{m}{m + 1}$ and $	an \beta = \frac{1}{2m + 1}$.Using the formula $	an(\alpha + \beta) = \frac{	an \alpha + 	an \be

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$\frac{\pi}{4}$

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(B) We have $	an \alpha = \frac{m}{m + 1}$ and $	an \beta = \frac{1}{2m + 1}$.Using the formula $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$:$	an(\alpha + \beta) = \frac{\frac{m}{m + 1} + \frac{1}{2m + 1}}{1 - \left(\frac{m}{m + 1}ight) \left(\frac{1}{2m + 1}ight)}$$= \frac{\frac{m(2m + 1) + (m + 1)}{(m + 1)(2m + 1)}}{\frac{(m + 1)(2m + 1) - m}{(m + 1)(2m + 1)}}$$= \frac{2m^2 + m + m + 1}{2m^2 + m + 2m + 1 - m}$$= \frac{2m^2 + 2m + 1}{2m^2 + 2m + 1} = 1$Since $	an(\alpha + \beta) = 1$,we have $\alpha + \beta = \frac{\pi}{4}$.

If $\tan \alpha = \frac{m}{m + 1}$ and $\tan \beta = \frac{1}{2m + 1}$,then $\alpha + \beta = $

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