Given that `tanalpha=m//(m+1),tanbeta=1//(2m+1)`, then what is the value of `alpha+beta` ?

Options

A. 0
B. `(pi)/(4)`
C. `(pi)/(6)`
D. `(pi)/(3)`

Answer

B

Explanation

As given , tan `alpha=(m)/(m+1)andtanbeta=(1)/(2m+1)`
`tan(alpha+beta)=(tanalpha+tanbeta)/(1-tan alphatanbeta)`
`=((m)/(m+1)+(1)/(2m+1))/(1-(m)/(m+1)xx(1)/(2m+1))=(m(2m+1)+(m+1))/((m+1)(2m+1)-m)`
`=(2m^(2)+2m+1)/(2m^(2)+2m+1)=1` ltbrlt So , `alpha+beta=(pi)/(4)`