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
