<p>If in a triangle <i>ABC</i>, cot <frac><i>A</i>}{2} + cot <frac><i>B</i>}{2} + cot <frac><i>C</i>}{2} = <i>X</i> cot <frac><i>A</i>}{2} cot <frac><i>B</i>}{2} cot <frac><i>C</i>}{2}, then find the value of <i>X</i>.</p>
Step-by-Step Solution
Key Concept: Use the fundamental identity for half-angles in a triangle: A + B + C = π, which leads to a relationship between cot(A/2), cot(B/2), and cot(C/2). The key is expressing tan((A+B)/2) in terms of these cotangents and using the constraint that A + B = π - C.
<p><strong>Step 1:</strong> Since A + B + C = π in any triangle, we have A + B = π - C.</p><p><strong>Step 2:</strong> Divide by 2: (A+B)/2 = π/2 - C/2.</p><p><strong>Step 3:</strong> Take tangent of both sides: tan((A+B)/2) = tan(π/2 - C/2) = cot(C/2).</p><p><strong>Step 4:</strong> Using the tangent addition formula: tan((A+B)/2) = (tan(A/2) + tan(B/2))/(1 - tan(A/2)tan(B/2)).</p><p><strong>Step 5:</strong> Convert to cotangent form. Let cot(A/2) = p, cot(B/2) = q, cot(C/2) = r. Then tan(A/2) = 1/p, tan(B/2) = 1/q.</p><p><strong>Step 6:</strong> From Step 3 and Step 4: (1/p + 1/q)/(1 - 1/(pq)) = r.</p><p><strong>Step 7:</strong> Simplify the left side: ((p+q)/(pq))/((pq-1)/(pq)) = (p+q)/(pq-1) = r.</p><p><strong>Step 8:</strong> Cross multiply: p + q = r(pq - 1) = rpq - r.</p><p><strong>Step 9:</strong> Rearrange: p + q + r = rpq.</p><p><strong>Step 10:</strong> Divide both sides by pqr: (p+q+r)/(pqr) = 1, which means p/pqr + q/pqr + r/pqr = 1.</p><p><strong>Step 11:</strong> This gives us: cot(A/2) + cot(B/2) + cot(C/2) = 1 · cot(A/2)cot(B/2)cot(C/2).</p><p><strong>∴ Answer:</strong> 1</p>
Correct Answer: 1