Question

In a triangle \( ABC \) with usual notations, if \[ \tan \left( \frac{B-C}{2} \right) = x \cot \left( \frac{A}{2} \right), \] then \( x = \, ? \) 

• A. \(\frac{c-a}{c+a}\)
• B. \(\frac{a-b}{a+b}\)
• C. \(\frac{b-c}{b+c}\)
• D. \(\frac{a+b}{a-b}\)

Hint:

The formula \[ \tan\left(\frac{B-C}{2}\right)=\frac{b-c}{b+c}\cot\left(\frac{A}{2}\right) \] is a standard triangle identity and is worth memorizing directly.

Answer 1

The Correct Option is C

Concept:
In a triangle, a standard identity is: \[ \tan\left(\frac{B-C}{2}\right)=\frac{b-c}{b+c}\cot\left(\frac{A}{2}\right) \] ip

Step 1: Compare the given relation with the standard identity.
Given: \[ \tan\left(\frac{B-C}{2}\right)=x\cot\left(\frac{A}{2}\right) \] Standard identity says: \[ \tan\left(\frac{B-C}{2}\right)=\frac{b-c}{b+c}\cot\left(\frac{A}{2}\right) \] ip

Step 2: Match the coefficients.
Comparing both expressions, we get: \[ x=\frac{b-c}{b+c} \] ip Hence, the correct answer is:
\[ \boxed{(C)\ \frac{b-c}{b+c}} \]