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

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(C) We use the formula for the tangent of a sum of two angles: $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$.Su

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$\frac{\pi}{4}$

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(C) We use the formula for the tangent of a sum of two angles: $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$.Substituting the given values:$	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)}$Simplifying the numerator: $\frac{m(2m+1) + 1(m+1)}{(m+1)(2m+1)} = \frac{2m^2 + m + m + 1}{(m+1)(2m+1)} = \frac{2m^2 + 2m + 1}{(m+1)(2m+1)}$.Simplifying the denominator: $1 - \frac{m}{(m+1)(2m+1)} = \frac{(m+1)(2m+1) - m}{(m+1)(2m+1)} = \frac{2m^2 + 3m + 1 - m}{(m+1)(2m+1)} = \frac{2m^2 + 2m + 1}{(m+1)(2m+1)}$.Dividing the numerator by the denominator:$	an(\alpha + \beta) = \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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