`sin alpha+ sinbeta=(1)/(4)` and `cos alpha+cos beta=(1)/(3)`
The value of `tan (alpha+beta)` is

Options

A. `(25)/(7)`
B. `(25)/(12)`
C. `(25)/(13)`
D. `(24)/(7)`

Answer

D

Explanation

`sin alpha + sin beta = (1)/(4)" "`...(i)
`rArr 2 sin((alpha + beta)/(2)) cos ((alpha -beta)/(2))= (1)/(4)" "`...(ii)
`rArr cosalpha + cos beta = (1)/(3)" " `...(iii)
`rArr 2 cos ((alpha + beta )/(2) ) cos((alpha -beta) /(2)) = (1)/(3)" "` (iv)
Dividing Eq. (ii) by Eq. (iv), we have
`" " tan((alpha + beta) /(2)) = (3)/(4)`
`rArr sin(alpha + beta) = (2tan((alpha + beta)/(2)))/(1+tan^(2)((alpha + beta)/(2)))`
`" " = (2xx (3)/(4))/(1+((3)/(4))^(2)) = (24)/(25)`
`therfore cos (alpha + beta) = (7)/(25) and tan(alpha + beta) = (24)/(7)`