Question

In a triangle ABC, with usual notations, $\cot\left(\frac{A+B}{2}\right) \cdot \tan\left(\frac{A-B}{2}\right) = $ ______.

• A. $\frac{a+b}{a-b}$
• B. $\frac{a-b}{a+b}$
• C. $\frac{a}{a+b}$
• D. $\frac{b}{a-b}$

Hint:

This derivation is essentially the proof of Napier's Analogy (The Law of Tangents)!
$\tan\left(\frac{A-B}{2}\right) = \frac{a-b}{a+b} \cot\left(\frac{C}{2}\right)$. Since $\cot(C/2) = \tan(\frac{A+B}{2})$, moving it over gives exactly this result.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem relates to Napier's Analogy (Tangent Rule) in trigonometry.

Step 2: Formula Application:
Napier's Analogy states: $\tan\left(\frac{A-B}{2}\right) = \frac{a-b}{a+b} \cot\left(\frac{C}{2}\right)$.

Step 3: Explanation:
In a triangle, $A+B+C = \pi$, so $\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}$. Thus, $\cot\left(\frac{A+B}{2}\right) = \cot\left(\frac{\pi}{2} - \frac{C}{2}\right) = \tan\left(\frac{C}{2}\right)$. The expression becomes: $\tan\left(\frac{C}{2}\right) \cdot \tan\left(\frac{A-B}{2}\right)$. Substitute Napier's Analogy: $\tan\left(\frac{C}{2}\right) \cdot \left[ \frac{a-b}{a+b} \cot\left(\frac{C}{2}\right) \right]$. Since $\tan \theta \cdot \cot \theta = 1$, the expression simplifies to $\frac{a-b}{a+b}$.

Step 4: Final Answer:
The value is $\frac{a-b}{a+b}$.