Given
cosα+cosβ=23
sinα+sinβ=21
⇒2cos(2α+β)cos(2α−β)=23 ......(i) and
2sin(2α+β)cos(2α−β)=21 ....(ii)
Dividing (ii) and (i)
tan(2(α+β))
∴tanθ=31
sin2θ+cos2θ= 1+tan2θ2tanθ+1+tan2θ1−tan2θ=1+912(31)+1+911−91
=91032+98=1014=57