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Question

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

A
10
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B
10
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C
25
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D
25
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Solution

The correct option is B 25

Using the fact that tanA and tanB are the roots of 3x210x25=0,
Sum of the roots =103 Product of the roots =253
tan(A+B)=tanA+tanB1tanAtanB
we get , tan(A+B)=10/328/3=514

We have,
cos2(A+B)=1+2cos2(A+B)=1tan2(A+B)1+tan2(A+B)=1251961+25196cos2(A+B)=196221

We see that
3sin2(A+B)10sin(A+B)cos(A+B)25cos2(A+B)

=cos2(A+B)(3tan2(A+B)10tan(A+B)25)

=757004900196×121221

=5525196×196221=25

So option C is the correct answer.

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