If `cosA+cosB=m` and `sinA+sinB=n` then `sin(A+B)=`

Options

A. `(mn)/(m^(2)+n^(2))`
B. `(2mn)/(m^(2)+n^(2))`
C. `(m^(2)+n^(2))/(2mn)`
D. `(mn)/(m+n)`

Answer

B

Explanation

Let `cosA+cosB=m` . . . (i)
and sinA+sin+B=n . . . (ii)
Consider sin (A+B) = `((m^(2)+n^(2))sin(A+B))/(m^(2)+n^(2))`
`=([2+2cos(A-B)]sin(A+B))/(2+2cos(A-B))` (from i and ii)
`(2sin(A+B)=2sin(A+B)cos(A-B))/(1+1+2cos(A-B))`
`=(2sin(A+B)+sin(A+B+A-B)+sin(A+B-A+B))/(1+1+2cos(A-B))`
`=(2sin(A+B)+sin2A+sin2B)/(1+1+2cos(A-B))`
`=(2(cosA+cosB)(sinA+sinB))/(sin^(2)A+cos^(2)A+sin^(2)B+cos^(2)B+2cosAcosB+2sinAsinB)`
`=(2(cosA+cosB)(sinA+sinB))/((sinA+sinB)^(2)+(cosA+cosB)^(2))`
`=(2mm)/(m^(2)+n)` (from (i) and (ii))
Hence , sin (A+B) `=(2mm)/(m^(2)+n^(2))`