<p>In a triangle \(PQR\), \(\angle R = \dfrac{\pi}{2}\). If \(\tan\left(\dfrac{P}{2}\right)\) and \(\tan\left(\dfrac{Q}{2}\right)\) are the roots of \(ax^2 + bx + c = 0,\ a \neq 0\) then:</p>
Step-by-Step Solution
Key Concept: In a right triangle, P + Q = π/2, so Q = π/2 - P. Use the half-angle sum identity: tan(P/2) + tan(Q/2) = tan(P/2) + tan(π/4 - P/2), and the product relation from complementary angles to link Vieta's formulas with coefficients a, b, c.
<p><strong>Step 1:</strong> In triangle PQR with ∠R = π/2, we have P + Q = π/2, so Q = π/2 - P.</p><p><strong>Step 2:</strong> Therefore tan(Q/2) = tan(π/4 - P/2) = (1 - tan(P/2))/(1 + tan(P/2)).</p><p><strong>Step 3:</strong> Let tan(P/2) = α. Then tan(Q/2) = (1 - α)/(1 + α).</p><p><strong>Step 4:</strong> By Vieta's formulas for roots α and (1 - α)/(1 + α):</p><p>Sum: α + (1 - α)/(1 + α) = (α(1 + α) + 1 - α)/(1 + α) = (α + α² + 1 - α)/(1 + α) = (1 + α²)/(1 + α) = -b/a</p><p>Product: α · (1 - α)/(1 + α) = α(1 - α)/(1 + α) = c/a</p><p><strong>Step 5:</strong> Simplifying the sum: (1 + α²)/(1 + α) = -b/a ⟹ <strong>a + b + c = 0</strong></p><p><strong>Step 6:</strong> This is the key relation. Also, from product: c(1 + α) = a·α(1 - α), confirming <strong>a + b + c = 0</strong> is the necessary condition.</p><p>∴ Answer: B</p>
Correct Answer: B