Law of Tangents/Corollary

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Corollary to Law of Tangents

Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.

Then:

$\tan \dfrac {A - B} 2 = \dfrac {a - b} {a + b} \cot \dfrac C 2$


Proof 1

\(\ds \dfrac {a + b} {a - b}\) \(=\) \(\ds \dfrac {\tan \frac 1 2 \paren {A + B} } {\tan \frac 1 2 \paren {A - B} }\) Law of Tangents
\(\ds \leadsto \ \ \) \(\ds \tan \frac {A - B} 2\) \(=\) \(\ds \dfrac {a - b} {a + b} \tan \frac {A + B} 2\) algebraic manipulation
\(\ds \) \(=\) \(\ds \dfrac {a - b} {a + b} \tan \frac {180 \degrees - C} 2\) Sum of Angles of Triangle equals Two Right Angles
\(\ds \) \(=\) \(\ds \dfrac {a - b} {a + b} \map \tan {90 \degrees - \frac C 2}\)
\(\ds \) \(=\) \(\ds \dfrac {a - b} {a + b} \cot \dfrac C 2\) Tangent of Complement equals Cotangent

$\blacksquare$


Proof 2

\(\ds \dfrac {a - b} {a + b}\) \(=\) \(\ds \dfrac {2 R \sin A - 2 R \sin B} {2 R \sin A + 2 R \sin B}\) Law of Sines
\(\ds \) \(=\) \(\ds \dfrac {2 \cos \frac {A + B} 2 \sin \frac {A - B} 2} {2 \sin \frac {A + B} 2 \cos \frac {A - B} 2}\) Sine minus Sine, Sine plus Sine
\(\ds \) \(=\) \(\ds \dfrac {\tan \frac {A - B} 2} {\tan \frac {A + B} 2}\) Tangent is Sine divided by Cosine and simplification
\(\ds \) \(=\) \(\ds \dfrac {\tan \frac {A - B} 2} {\tan \frac {180 \degrees - C} 2}\) Sum of Angles of Triangle equals Two Right Angles
\(\ds \) \(=\) \(\ds \dfrac {\tan \frac {A - B} 2} {\map \tan {90^\circ - \frac C 2} }\) simplification
\(\ds \) \(=\) \(\ds \dfrac {\tan \frac {A - B} 2} {\cot \dfrac C 2}\) Tangent of Complement equals Cotangent
\(\ds \leadsto \ \ \) \(\ds \dfrac {a - b} {a + b} \cot \dfrac C 2\) \(=\) \(\ds \tan \frac {A - B} 2\) Tangent of Complement equals Cotangent

Hence the result.

$\blacksquare$


Also presented as

The Corollary to the Law of Tangents can also be found presented as:

$\dfrac {a - b} {a + b} = \tan \dfrac {A - B} 2 \tan \dfrac C 2$


Also known as

Some sources present this as the Law of Tangents itself.


Examples

Arbitrary Example $1$

Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.

Let $\triangle ABC$ be such that:

\(\ds a\) \(=\) \(\ds 18.4\)
\(\ds b\) \(=\) \(\ds 12.2\)
\(\ds C\) \(=\) \(\ds 42 \degrees\)

where the units of measurement of $a$ and $b$ are arbitrary.


Then:

\(\ds A\) \(=\) \(\ds 96 \degrees 50'\)
\(\ds B\) \(=\) \(\ds 41 \degrees 10'\)
\(\ds c\) \(=\) \(\ds 12.4\)


Sources

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