Given equation is x2−(sinα−2)x−(1+sinα)=0 Let x1 and x2 be two roots of quadratic equation ∴x1+x2=sinα−2 and x1x2=−(1+sinα) (x1+x2)2=(sinα−2)2=sin2α+4−4sinα ⇒x12+x22=sin2α+4−4sinα−2x1x2 =sin2α+4−4sinα+2(1+sinα) =sin2α−2sinα+6 Now, By putting α=
π
6
,α=
π
4
,α=
π
3
and α=
π
2
in (i) one by one We get least value of x12+x22 at