Q.Prove that $\tan\left(\dfrac{B-C}{2}\right)=\dfrac{b-c}{b+c}\cot\dfrac{A}{2}$, and verify it for the triangle with $a=13,\,b=14,\,c=15$.

Tangent Rule (Napier's Analogy)

When a triangle is given by ASA or AAS data — two angles and a side — the sine rule alone can find the remaining sides once the third angle is known. But sometimes a problem is phrased the other way: you know a side-difference like $b-c$, and want an angle-difference like $B-C$. Napier's Analogy (the tangent rule) bridges exactly this gap:

and cyclically. Historically (before calculators), this identity mattered enormously: it let a navigator or surveyor compute an angle using only logarithm tables of tangents and cotangents — no need to look up an inverse sine or cosine of a messy fraction.

Where it comes from

It follows from the sine rule plus the sum-to-product identities. Since $b=2R\sin B$ and $c=2R\sin C$, the ratio $\frac{b-c}{b+c}$ becomes $\frac{\sin B-\sin C}{\sin B+\sin C}$, which sum-to-product turns into $\cot\frac{B+C}2\tan\frac{B-C}2$. Because $\frac{B+C}2=90^\circ-\frac A2$, the cotangent term simplifies to $\tan\frac A2$, and rearranging gives the stated formula.

How it is used today

Even without needing log tables, Napier's analogy is still the standard technique for identity-proving exercises that link a half-angle-difference to a side ratio, and it appears whenever a "solve the triangle" problem is more naturally phrased through $B-C$ than through $B$ and $C$ separately. It is also the cleanest packaging of a sum-to-product manipulation into one memorable formula.

A common trap …