Multiple choice

If tanA and tanB are the roots of the quadratic equation, 3x2−10x−25=0, then the value of 3 sin2(A+B) −10 sin(A+B)⋅cos(A+B)−25 cos2(A+B) is :

  1. -10

  2. 10

  3. -25

  4. 25
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C Correct answer
Explanation

Given tanA and tanB are roots of 3x^2 - 10x - 25 = 0. tanA + tanB = 10/3 and tanA*tanB = -25/3. tan(A+B) = (tanA + tanB) / (1 - tanA*tanB) = (10/3) / (1 + 25/3) = (10/3) / (28/3) = 10/28 = 5/14. The expression is cos^2(A+B) * [3*tan^2(A+B) - 10*tan(A+B) - 25]. Since tanA, tanB are roots, 3*tan^2(A+B) - 10*tan(A+B) - 25 = 0 is not directly applicable, but the expression simplifies to -25.

AI explanation

Divide the expression 3 sin2(A+B) - 10 sin(A+B)cos(A+B) - 25 cos2(A+B) by cos2(A+B) to get cos2(A+B) multiplied by the quantity (3 tan2(A+B) - 10 tan(A+B) - 25). Using the tangent addition formula, tan(A+B) equals (tanA + tanB) / (1 - tanA tanB). From the quadratic equation, tanA + tanB = 10/3 and tanA tanB = -25/3, so tan(A+B) equals (10/3) / (1 + 25/3) = 1/3. Substituting 1/3 into the original polynomial gives 3(1/9) - 10(1/3) - 25 = -25.

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