Tan 3A Formula – Proof and Examples

Here you will learn what is the formula of tan 3A in terms of A with proof and examples based on it.

Let’s begin –

Tan 3A Formula

The formula of tan 3A is \(3 tan A – tan^3 A\over 1 – 3 tan^2 A\).

Proof :

We have,

tan (A + B) = \(tan A + tan B\over 1 – tan A tan B\)

Replacing B by 2A,

\(\implies\) tan 3A = \(tan A + tan 2A\over 1 – tan A tan 2A\)

Since, tan 2A = \(2 tan A\over tan^2 A\)

\(\implies\) tan 3A = \(tan A + {2 tan A\over tan^2 A}\over 1 – tan A \times {2 tan A\over tan^2 A}\)

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

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

We can also write above relation of angle A in terms of angle A/3, just replace A by A/3, we get

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

Example : 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

Leave a Comment

Your email address will not be published.