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If tanalpha=m/(m+1) and tanbeta=1/(2m+1)...

If `tanalpha=m/(m+1)` and `tanbeta=1/(2m+1)`. Find the possible values of `tan(alpha+beta)`

A

`2`

B

`1`

C

`-1`

D

0

Text Solution

Verified by Experts

The correct Answer is:
B
We have `tan (alpha+beta)=(tan alpha+tan beta)/(1-tan alpha tan beta)`
`=((m)/(m+1)+(1)/(2m+1))/(1-(m)/(m+1)xx(1)/(2m+1))`
`=(2m^(2)+2m+1)/(2m^(2)+2m+1)=1`
`rArr (alpha+beta)_("least")=(pi)/(4)`.
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