If cosα+cosβ=0=sinα+sinβ then cos2α+cos2β is equal to
2cos(α+β)
-2cos(α+β)
3cos(α+β)
None of these
Given cosα+cosβ=0 and sinα+sinβ=0
On squaring and subtracting both the equations, we get
(cosα+cosβ)2−(sinα+sinβ)2=0⇒cos2α+cos2β+2cosαcosβ−sin2α+sin2β+2sinαsinβ=0⇒cos2α−sin2α+cos2β−sin2β+2[cosαcosβ−sinαsinβ]=0⇒cos2α+cos2β+2cos(α+β)=0⇒cos2α+cos2β=−2cos(α+β)