<p>If in a triangle <i>ABC</i>, <frac>sin <i>A</i> + sin <i>B</i> + sin <i>C</i>}{sin <i>A</i> + sin <i>B</i> - sin <i>C</i>} = 2<i>X</i> cot <frac><i>A</i>}{2} cot <frac><i>B</i>}{2}, then find the value of <i>X</i>.</p>
Step-by-Step Solution
Key Concept: Use the sine rule to convert sines to sides, then apply Mollweide's formula and half-angle tangent formulas to relate the given ratio to cotangent terms.
**Step 1:** Apply the sine rule.
In any triangle ABC, the sine rule states $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$, where $R$ is the circumradius.
Thus, $\sin A = \frac{a}{2R}$, $\sin B = \frac{b}{2R}$, and $\sin C = \frac{c}{2R}$.
**Step 2:** Substitute these into the left-hand side of the given expression.
$$ \frac{\sin A + \sin B + \sin C}{\sin A + \sin B - \sin C} = \frac{\frac{a}{2R} + \frac{b}{2R} + \frac{c}{2R}}{\frac{a}{2R} + \frac{b}{2R} - \frac{c}{2R}} = \frac{a + b + c}{a + b - c} $$
**Step 3:** Rewrite the expression using Mollweide's formula.
The expression $\frac{a+b+c}{a+b-c}$ can be written as $\frac{\frac{a+b}{c} + 1}{\frac{a+b}{c} - 1}$.
Mollweide's formula states $\frac{a+b}{c} = \frac{\cos\left(\frac{A-B}{2}\right)}{\sin\left(\frac{C}{2}\right)}$.
Substituting this into the expression:
$$ \frac{a + b + c}{a + b - c} = \frac{\frac{\cos\left(\frac{A-B}{2}\right)}{\sin\left(\frac{C}{2}\right)} + 1}{\frac{\cos\left(\frac{A-B}{2}\right)}{\sin\left(\frac{C}{2}\right)} - 1} = \frac{\cos\left(\frac{A-B}{2}\right) + \sin\left(\frac{C}{2}\right)}{\cos\left(\frac{A-B}{2}\right) - \sin\left(\frac{C}{2}\right)} $$
**Step 4:** Apply angle sum identity and sum-to-product formulas.
In a triangle, $A+B+C = \pi$, so $\frac{C}{2} = \frac{\pi}{2} - \frac{A+B}{2}$.
Therefore, $\sin\left(\frac{C}{2}\right) = \sin\left(\frac{\pi}{2} - \frac{A+B}{2}\right) = \cos\left(\frac{A+B}{2}\right)$.
Substitute this into the expression from Step 3:
$$ \frac{\cos\left(\frac{A-B}{2}\right) + \cos\left(\frac{A+B}{2}\right)}{\cos\left(\frac{A-B}{2}\right) - \cos\left(\frac{A+B}{2}\right)} $$
Using the sum-to-product identities:
$\cos X + \cos Y = 2\cos\left(\frac{X+Y}{2}\right)\cos\left(\frac{X-Y}{2}\right)$
$\cos X - \cos Y = -2\sin\left(\frac{X+Y}{2}\right)\sin\left(\frac{X-Y}{2}\right)$
Let $X = \frac{A-B}{2}$ and $Y = \frac{A+B}{2}$.
Then $\frac{X+Y}{2} = \frac{A}{2}$ and $\frac{X-Y}{2} = \frac{-B}{2}$.
The numerator becomes $2\cos\left(\frac{A}{2}\right)\cos\left(\frac{-B}{2}\right) = 2\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)$.
The denominator becomes $-2\sin\left(\frac{A}{2}\right)\sin\left(\frac{-B}{2}\right) = 2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)$.
Thus, the expression simplifies to:
$$ \frac{2\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)}{2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)} = \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} \cdot \frac{\cos\left(\frac{B}{2}\right)}{\sin\left(\frac{B}{2}\right)} = \cot\left(\frac{A}{2}\right)\cot\left(\frac{B}{2}\right) $$
**Step 5:** Compare with the given right-hand side.
The problem states that
$$ \frac{\sin A + \sin B + \sin C}{\sin A + \sin B - \sin C} = X \cot\left(\frac{A}{2}\right)\cot\left(\frac{B}{2}\right) $$
From Step 4, the left-hand side is $\cot\left(\frac{A}{2}\right)\cot\left(\frac{B}{2}\right)$.
Therefore, we have:
$$ \cot\left(\frac{A}{2}\right)\cot\left(\frac{B}{2}\right) = X \cot\left(\frac{A}{2}\right)\cot\left(\frac{B}{2}\right) $$
Assuming $\cot\left(\frac{A}{2}\right)\cot\left(\frac{B}{2}\right) \neq 0$, we can divide both sides by this term:
$$ 1 = X $$
Thus, the value of $X$ is $1$.
Correct Answer: 1