Prove that tan A + tan (60 + A) –  tan (60 – A) = 3 tan 3A.

Solution :

We have,

L.H.S = tan A + tan (60 + A) –  tan (60 – A)

\(\implies\) L.H.S = tan A + \(\sqrt{3} + tan A\over 1 – \sqrt{3} tan A\) – \(\sqrt{3} – tan A\over 1 + \sqrt{3} tan A\)

[ By using this formula, tan (A + B) = \(tan A + tan B\over 1 – tan A tan B\)  above ]

\(\implies\) L.H.S = tan A + \(8 tan A\over 1 – 3 tan^2 A\)

\(\implies\) L.H.S = \(9 tan A – 3 tan^3 A\over 1 – 3 tan^2 A\)

L.H.S = 3(\(3 tan A – tan^3 A\over 1 – 3 tan^2 A\))

Since \(3 tan A – tan^3 A\over 1 – 3 tan^2 A\) = tan 3A

\(\implies\) L.H.S = 3 tan 3A = R.H.S

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