Properties of Triangles – Tangent Rule and Area of Triangle
Tangent rule (or) Napier’s analogy:
In ΔABC,
\(\tan \left( \frac{A-B}{2} \right)=\left( \frac{a-b}{a+b} \right)\cot \left( \frac{C}{2} \right)\).
\(\tan \left( \frac{B-C}{2} \right)=\left( \frac{b-c}{b+c} \right)\cot \left( \frac{A}{2} \right)\).
\(\tan \left( \frac{C-A}{2} \right)=\left( \frac{c-a}{c+a} \right)\cot \left( \frac{B}{2} \right)\).
Area of ΔABC is given by:
(I) Δ = ½ ab sin C; Δ = ½ bc sin A; Δ = ½ ca sin B,
(II) \(\Delta =\sqrt{s(s-a)(s-b)(s-c)}\),
(III) \(\Delta =\frac{abc}{4R}\),
(IV) Δ = 2R² sin A sin B sin C.
(V) Δ = rs
(VI) \(\Delta =\sqrt{r{{r}_{1}}{{r}_{2}}{{r}_{3}}}\).
Example: Prove that \(\left( \frac{1}{r}-\frac{1}{{{r}_{1}}} \right)\left( \frac{1}{r}-\frac{1}{{{r}_{2}}} \right)\left( \frac{1}{r}-\frac{1}{{{r}_{3}}} \right)=\frac{abc}{{{\Delta }^{3}}}=\frac{4R}{{{r}^{2}}{{S}^{2}}}\).
Solution: Given that,
\(\left( \frac{1}{r}-\frac{1}{{{r}_{1}}} \right)\left( \frac{1}{r}-\frac{1}{{{r}_{2}}} \right)\left( \frac{1}{r}-\frac{1}{{{r}_{3}}} \right)=\frac{abc}{{{\Delta }^{3}}}=\frac{4R}{{{r}^{2}}{{S}^{2}}}\),
\(\left( \frac{1}{r}-\frac{1}{{{r}_{1}}} \right)=\frac{S}{\Delta }-\frac{S-a}{\Delta }\),
\(\left( \frac{1}{r}-\frac{1}{{{r}_{1}}} \right)=\frac{S-S+a}{\Delta }\),
\(\left( \frac{1}{r}-\frac{1}{{{r}_{1}}} \right)=\frac{a}{\Delta }\),
\(\left( \frac{1}{r}-\frac{1}{{{r}_{2}}} \right)=\frac{S}{\Delta }-\frac{S-b}{\Delta }\),
\(\left( \frac{1}{r}-\frac{1}{{{r}_{2}}} \right)=\frac{S-S+b}{\Delta }\),
\(\left( \frac{1}{r}-\frac{1}{{{r}_{1}}} \right)=\frac{b}{\Delta }\),
\(\left( \frac{1}{r}-\frac{1}{{{r}_{3}}} \right)=\frac{S}{\Delta }-\frac{S-c}{\Delta }\),
\(\left( \frac{1}{r}-\frac{1}{{{r}_{3}}} \right)=\frac{S-S+c}{\Delta }\),
\(\left( \frac{1}{r}-\frac{1}{{{r}_{3}}} \right)=\frac{c}{\Delta }\),
L.H.S \(\left( \frac{1}{r}-\frac{1}{{{r}_{1}}} \right)\left( \frac{1}{r}-\frac{1}{{{r}_{2}}} \right)\left( \frac{1}{r}-\frac{1}{{{r}_{3}}} \right)\),
\(=\frac{a}{\Delta }\times \frac{b}{\Delta }\times \frac{c}{\Delta }\),
\(=\frac{abc}{{{\Delta }^{3}}}\),
\(=\frac{4R\times \Delta }{{{\Delta }^{3}}}\),
\(=\frac{4R}{{{\Delta }^{2}}}\),
\(=\frac{4R}{{{\left( rS \right)}^{2}}}\) = R.H.S.
Hence proved \(\left( \frac{1}{r}-\frac{1}{{{r}_{1}}} \right)\left( \frac{1}{r}-\frac{1}{{{r}_{2}}} \right)\left( \frac{1}{r}-\frac{1}{{{r}_{3}}} \right)=\frac{abc}{{{\Delta }^{3}}}=\frac{4R}{{{r}^{2}}{{S}^{2}}}\).