Question:

<p><strong>Step 1:</strong> Given that sin A, sin B, sin C are in AP, we have:</p><p>2sin B = sin A + sin C</p><p><strong>Step 2:</strong> By the sine rule: a/sin A = b/sin B = c/sin C = 2R</p><p>Therefore: sin A = a/(2R), sin B = b/(2R), sin C = c/(2R)</p><p>Substituting into the AP condition:</p><p>2b = a + c</p><p><strong>Step 3:</strong> Now express tan(B/2) using the half-angle formula:</p><p>tan(B/2) = r/(s-b), where r is inradius and s is semi-perimeter</p><p>Also, tan(B/2) = (s-a)(s-c)/(s(s-b)) using another formula</p><p><strong>Step 4:</strong> From 2b = a + c, we have s = (a+b+c)/2 = (2b+b)/2 = 3b/2</p><p>Therefore: s - b = b/2, s - a = (3b/2 - a), s - c = (3b/2 - c)</p><p><strong>Step 5:</strong> Using tan(B/2) = √[(s-a)(s-c)]/[s(s-b)] with constraint 2b = a + c:</p><p>Let a + c = 2b. By AM-GM inequality: (a+c)/2 ≥ √(ac), so b ≥ √(ac)</p><p>This gives ac ≤ b²</p><p><strong>Step 6:</strong> tan(B/2) = √[(s-a)(s-c)]/[s(s-b)]</p><p>With s = 3b/2: tan(B/2) = √[(3b/2-a)(3b/2-c)]/[(3b/2)(b/2)]</p><p>Let a + c = 2b. Then (3b/2 - a) + (3b/2 - c) = 3b - 2b = b</p><p><strong>Step 7:</strong> For fixed sum (3b/2-a) + (3b/2-c) = b, the product (3b/2-a)(3b/2-c) is maximized when 3b/2-a = 3b/2-c, which means a = c</p><p>Maximum product = (b/2)² = b²/4</p><p><strong>Step 8:</strong> Therefore: tan(B/2)_max = √(b²/4)/[(3b/2)(b/2)] = (b/2)/(3b²/4) = (b/2) × (4/3b²) = 2/(3b)</p><p>When a = c and 2b = a + c = 2a, we get b = a, so triangle is equilateral with B = 60°</p><p>tan(30°) = 1/√3</p><p><strong>∴ Answer:</strong> A</p>