If cos⁡α+cos⁡β=0=sin⁡α+sin⁡β then cos⁡2α+cos⁡2β is equal to

Given cos⁡α+cos⁡β=0 and sin⁡α+sin⁡β=0On squaring and subtracting both the equations, we get (cos⁡α+cos⁡β)2−(sin⁡α+sin⁡β)2=0⇒cos2⁡α+cos2⁡β+2cos⁡αcos⁡β−sin2⁡α+sin2⁡β+2sin⁡αsin⁡β=0⇒cos2⁡α−sin2⁡α+cos2⁡β−sin2⁡β+2[cos⁡αcos⁡β−sin⁡αsin⁡β]=0⇒cos⁡2α+cos⁡2β+2cos⁡(α+β)=0⇒cos⁡2α+cos⁡2β=−2cos⁡(α+β)