Napier's Analogy (Law of Tangents)
Category : JEE Main & Advanced
For any triangle ABC,
(1) \[\tan \left( \frac{A-B}{2} \right)=\left( \frac{a-b}{a+b} \right)\cot \frac{C}{2}\]
(2) \[\tan \left( \frac{B-C}{2} \right)=\left( \frac{b-c}{b+c} \right)\cot \frac{A}{2}\]
(3) \[\tan \left( \frac{C-A}{2} \right)=\left( \frac{c-a}{c+a} \right)\cot \frac{B}{2}\]
Mollweide's formula: For any triangle,
\[\frac{a+b}{c}=\frac{\cos \frac{1}{2}(A-B)}{\sin \frac{1}{2}C},\,\frac{a-b}{c}=\frac{\sin \frac{1}{2}(A-B)}{\cos \frac{1}{2}C}\].
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