Finding tan of difference of half angles (Napier Analogy)

\[\tan \frac{B-C}{2}=\frac{b-c}{b+c}\cot \frac{A}{2} \]

where B and C are the angles opposite to side AB and  BC  respectively  

Note       b is the side opposite to the angle B of the triangle

               c is the side opposite to the angle C of the triangle

Boundary condition b ,c must be positive

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