tan A+ tan (60+A) -tan (60-A) is equal to (1) 3 tan 3A (2) tan 3A (3) cot 3A (4) sin 3A

Answer: (1)

Let us calculate tan (60 + A) and tan (60 – A).

\(tan \left ( 60 + A \right ) = \frac{tan 60^{0}+ tan A}{1 – tan60^{0} tanA}\)

⇒ \(tan \left ( 60 + A \right ) = \frac{\sqrt{3}+tan A}{1 – \sqrt{3} tanA}\)…………………….(1)

\(tan \left ( 60 – A \right ) = \frac{tan 60^{0}- tan A}{1 + tan60^{0} tanA}\)

⇒ \(tan \left ( 60 – A \right ) = \frac{\sqrt{3} – tan A}{1 + \sqrt{3} tanA}\)……………………..(2

Subtracting equation (1) and (2)

\(tan \left ( 60 + A \right ) – tan \left ( 60 – A \right ) = \frac{\sqrt{3}+tan A}{1 – \sqrt{3} tanA} – \frac{\sqrt{3} – tan A}{1 + \sqrt{3} tanA}\) \(tan \left ( 60 + A \right ) – tan \left ( 60 – A \right ) = \frac{\left ( \sqrt{3} + tan A \right)\left ( 1+ \sqrt{3}tan A \right) – \left ( \sqrt{3} – tan A \right)\left ( 1- \sqrt{3}tan A \right )}{1^{2} – \left ( \sqrt{3} tan A \right )^{2}}\) \(tan \left ( 60 + A \right ) – tan \left ( 60 – A \right ) = \frac{\sqrt{3} + 3tan A + tan A + \sqrt{3} tan^{2}A – \sqrt{3} + 3 tan A + tan A – \sqrt{3} tan^{2}A}{1 – 3 tan^{2}A}\) \(tan \left ( 60 + A \right ) – tan \left ( 60 – A \right ) = \frac{8 tan A}{1 – 3 tan^{2}A}\) \(tan A + tan \left ( 60 + A \right ) – tan \left ( 60 – A \right ) = tan A + \frac{8 tan A}{1 – 3 tan^{2}A}\) \(tan A + tan \left ( 60 + A \right ) – tan \left ( 60 – A \right ) = \frac{tan A – 3 tan^{3}A + 8 tan A}{1 – 3 tan^{2}A}\) \(tan A + tan \left ( 60 + A \right ) – tan \left ( 60 – A \right ) = \frac{9 tan A – 3 tan^{2}A}{1 – 3 tan^{2}A}\) \(tan A + tan \left ( 60 + A \right ) – tan \left ( 60 – A \right ) = \frac{3\left ( 3 tan A – tan^{3}A \right )}{1 – 3 tan^{2}A}\) \(tan A + tan \left ( 60 + A \right ) – tan \left ( 60 – A \right ) = 3 tan 3A\)

Here tan 3A = \(tan 3A = \frac{3 tan A – tan^{3}A}{1 – 3 tan^{2}A}\)

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