If the points O(0,0), A(cosα, sinα),B(cosβ, sinβ) are the vertices of a right-angled triangle, then sinα−β2=
12
13
none of these
We have OA=OB=1
and AB=2−2cos(α−β)=2sinα−β2
Since ∆AOB is a right triangle. Therefore
AB=OA2+OB2⇒2sinα−β2=2⇒sinα−β2=12