# 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