<p>If \(\tan A\) and \(\tan B\) are the roots of the quadratic equation, \(3x^2-10x-25=0\), then the value of \(3\sin^2(A+B)-10\sin(A+B)\cos(A+B)-25\cos^2(A+B)\) is</p>
Step-by-Step Solution
Key Concept: Use Vieta's formulas to find tan A + tan B and tan A·tan B, then apply the tangent addition formula to find tan(A+B). Finally, divide the given expression by sin²(A+B) + cos²(A+B) = 1 to convert it into a polynomial in tan(A+B).
<p><strong>Step 1:</strong> From the quadratic equation 3x² - 10x - 25 = 0, by Vieta's formulas:</p><p>tan A + tan B = 10/3 and tan A · tan B = -25/3</p><p><strong>Step 2:</strong> Find tan(A+B) using the addition formula:</p><p>tan(A+B) = (tan A + tan B)/(1 - tan A tan B) = (10/3)/(1 - (-25/3)) = (10/3)/(28/3) = 10/28 = 5/14</p><p><strong>Step 3:</strong> Recognize that the expression 3sin²(A+B) - 10sin(A+B)cos(A+B) - 25cos²(A+B) can be written in terms of tan(A+B) by dividing numerator and denominator by cos²(A+B):</p><p>= cos²(A+B)[3tan²(A+B) - 10tan(A+B) - 25]</p><p><strong>Step 4:</strong> Since tan(A+B) = 5/14 is a root of 3x² - 10x - 25 = 0 (by construction), we have:</p><p>3(5/14)² - 10(5/14) - 25 = 3(25/196) - 50/14 - 25 = 75/196 - 700/196 - 4900/196 = -5525/196</p><p><strong>Step 5:</strong> We need cos²(A+B). Since tan(A+B) = 5/14, we have sec²(A+B) = 1 + tan²(A+B) = 1 + 25/196 = 221/196</p><p>Therefore, cos²(A+B) = 196/221</p><p><strong>Step 6:</strong> Final answer = (196/221) × (-5525/196) = -5525/221 = -25</p><p>∴ Answer: <strong>C (which equals -25)</strong></p>
Correct Answer: C