Solution :
We have,
L.H.S = cos A cos (60 – A) cos (60 + A)
\(\implies\) L.H.S = cos A (\(cos^2 60 – sin^2 A\))
[ By using this formula, cos (A + B) cos (A – B) = \(cos^2 A\) – \(sin^2 B\) above ]
\(\implies\) L.H.S = cos A (\(1\over 4\) – \(sin^2 A\)) = cos A \(({1\over 4} – (1 – cos^2 A))\)
\(\implies\) L.H.S = cos A (\({-3\over 4} + cos^2 A\))
\(\implies\) L.H.S = \(1\over 4\) cos A (\(-3 + 4 cos^2 A\)) = \(1\over 4\)(\(4 cos^3 A\) – 3 cos A)
Since \(4 cos^3 A\) – 3 cos A = cos 3A,
\(\implies\) L.H.S = \(1\over 4\) cos 3A = R.H.S
