If cosα + cosβ = 0 = sinα + sinβ,

Solution:

According to the question,

cosα + cosβ = 0 = sinα + sinβ …(i)

Since, LHS = cos 2α + cos 2β

We know that,

cos 2x = cos2x – sin2x

Therefore,

LHS = cos2α – sin2α + (cos2β – sin2β)

⇒ LHS = cos2α + cos2β – (sin2α + sin2β)

Also, since,

a2 + b2 = (a+b)2 – 2ab

⇒ LHS = (cosα + cosβ)2 – 2cosα cosβ –(sinα + sinβ)2 +2sinα sinβ

From equation (i),

⇒ LHS = 0 – 2cosα cosβ -0 + 2sinα sinβ

⇒ LHS = -2(cosα cosβ – sinα sinβ)

∵ cos (α + β) = cosα cosβ – sinα sinβ

Therefore, LHS = -2 cos (α + β) = RHS

Hence, cos 2α + cos 2β = –2cos (α + β)