If tan α=mm+1, tan β=12m+1 then α+β=

tan α=MM+1, tan β=12M+1 tan(α+β)=MM+1+12M+11-MM+1 12M+1 =2M2+2M+12M2+2M+1=1 α+β=π4

If tan α=mm+1, tan β=12m+1 then α+β=

tan α=MM+1, tan β=12M+1 tan(α+β)=MM+1+12M+11-MM+1 12M+1 =2M2+2M+12M2+2M+1=1 α+β=π4

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If $\tan α = \frac{m}{m + 1}, \tan β = \frac{1}{2 m + 1}$ then $α + β =$

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$nπ + \frac{π}{2}$

$nπ + \frac{π}{3}$

$nπ - \frac{π}{4}$

$nπ + \frac{π}{4}$

answer is D.

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$\tan α = \frac{M}{M + 1}, \tan β = \frac{1}{2 M + 1} \tan (α + β) = \frac{\frac{M}{M + 1} + \frac{1}{2 M + 1}}{1 - \frac{M}{M + 1} \frac{1}{2 M + 1}} = \frac{2 M^{2} + 2 M + 1}{2 M^{2} + 2 M + 1} = 1 α + β = \frac{π}{4}$

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If $\tan α = \frac{m}{m + 1}, \tan β = \frac{1}{2 m + 1}$ then $α + β =$