Cot 6° cot 42° CoT 66° cot78°= 1 Plzzzz give me answer

4 cosA.cos(60 - A).cos(60 + A) = 2 cos A{2 cos (60 + A) * cos (60 - A)} = 2 cos A{cos 120 + cos 2A} [ Applying the relation : 2 cos X * cos Y = cos (X + Y) + cos (X - Y) ] = 2cos A {- (1/2) + cos 2A} = - cos A + 2 cos 2A * cos A = - cos A + { cos (2A + A) + cos(2A - A) } [ Applying the same relation as before ] = - cos A + cos 3A + cos A = cos 3ASimilarly for 4cos A.cos 60+A.cos 60-A = cos 3A.Dividing both you get cotA.cot60-A.cot60+A = cot3A.Now it can be used twice in the question.Multiply and divide by cot18°.->[ (Cot18° cot78° cot42°) cot66°cot6° ] / cot18°->[ cot(3*18)° cot66° cot6° ] / cot18°->cot(3*6) / cot18°->cot18° / cot18° = 1Hence proved.