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 Question:
We must simplify a trigonometric expression involving half-angles and transform it into an expression containing only the side lengths ($a, b$) of the triangle using standard triangle laws.

Step 2: Detailed Explanation:
Let's break down the given expression by converting tangent and cotangent into sine and cosine:
Expression = $\cot\left(\frac{A+B}{2}\right) \cdot \tan\left(\frac{A-B}{2}\right)$
Expression = $\left[ \frac{\cos\left(\frac{A+B}{2}\right)}{\sin\left(\frac{A+B}{2}\right)} \right] \cdot \left[ \frac{\sin\left(\frac{A-B}{2}\right)}{\cos\left(\frac{A-B}{2}\right)} \right]$
Reorganize the terms to pair up the numerators and denominators:
Expression = $\frac{\cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)}{\sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)}$
Multiply the numerator and the denominator by 2 to perfectly match the standard product-to-sum trigonometric identities:
Expression = $\frac{2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)}{2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)}$
Apply the standard trigonometric identities:
$2 \cos X \sin Y = \sin(X + Y) - \sin(X - Y)$
$2 \sin X \cos Y = \sin(X + Y) + \sin(X - Y)$
Numerator becomes: $\sin(A) - \sin(B)$
Denominator becomes: $\sin(A) + \sin(B)$
So, the entire expression simplifies elegantly to:
Expression = $\frac{\sin A - \sin B}{\sin A + \sin B}$
Now, apply the fundamental Law of Sines ($\frac{a}{\sin A} = \frac{b}{\sin B} = 2R$, which means $\sin A \propto a$ and $\sin B \propto b$):
Expression = $\frac{a - b}{a + b}$

Step 3: Final Answer:
The expression is equal to $\frac{a-b}{a+b}$, matching option (b).