Law of Tangents
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Theorem
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:
- $\dfrac {a + b} {a - b} = \dfrac {\tan \frac 1 2 \paren {A + B} } {\tan \frac 1 2 \paren {A - B} }$
Corollary
- $\tan \dfrac {A - B} 2 = \dfrac {a - b} {a + b} \cot \dfrac C 2$
Proof
Let $d = \dfrac a {\sin A}$.
From the Law of Sines, let:
- $d = \dfrac a {\sin A} = \dfrac b {\sin B}$
so that:
\(\ds a\) | \(=\) | \(\ds d \sin A\) | |||||||||||||
\(\ds b\) | \(=\) | \(\ds d \sin B\) | |||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {a + b} {a - b}\) | \(=\) | \(\ds \frac {d \sin A + d \sin B} {d \sin A - d \sin B}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\sin A + \sin B} {\sin A - \sin B}\) | |||||||||||||
Let $C = \frac 1 2 \paren {A + B}$ and $D = \frac 1 2 \paren {A - B}$, and proceed as follows: | |||||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {a + b} {a - b}\) | \(=\) | \(\ds \frac {2 \sin C \cos D} {\sin A - \sin B}\) | Sine plus Sine | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 \sin C \cos D} {2 \sin D \cos C}\) | Sine minus Sine | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\frac {\sin C} {\cos C} } {\frac {\sin D} {\cos D} }\) | dividing top and bottom by $\cos C \cos D$ | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\tan C} {\tan D}\) | Tangent is Sine divided by Cosine | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\tan \frac 1 2 \paren {A + B} } {\tan \frac 1 2 \paren {A - B} }\) | substituting back for $C$ and $D$ |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.94$