If sinαsinβ−cosαcosβ+1=0 then 1+cotαtanβ=
-1
1
0
2
We have sinαsinβ-cosαcosβ=-1⇒cos(α+β)=1⇒α+β=0,2π,4π ……Now 1+cotαtanβ=1+cosαsinα·sinβcosβ =sin(α+β)sinαsinβ =0 ∵α+β=2nπ