Answer :

LHS = cos2A + cos2B – 2 cosA cosB cos(A +B)


= cos2A + 1 – sin2B - 2 cosA cosB cos(A +B)


= 1 + cos2A – sin2B - 2 cosA cosB cos(A +B)


We know that cos2A – sin2B = cos(A +B) cos(A –B)


= 1 + cos(A +B) cos(A –B) - 2 cosA cosB cos(A +B)


= 1 + cos(A +B) [cos(A –B) – 2 cosA cosB]


We know that cos(A -B) = cosA cosB + sinA sinB.


= 1 + cos(A +B) [cosA cosB + sinA sinB – 2 cosA cosB]


= 1 + cos(A +B) [-cosA cosB + sinA sinB]


= 1 - cos(A +B) [cosA cosB - sinA sinB]


We know that cos(A +B) = cosA cosB - sinA sinB.


= 1 – cos2(A +B)


= sin2(A +B) = RHS


Hence proved.


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