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If tanA and tanB are the roots of the quadratic equation, 3 x^{2}-10 x -25 =0, then the value of 3 \sin ^{2}(A +B) -10 \sin (A +B)\cdot \cos (A +B) -25 cos^{2}(A +B) is: 

  • Option 1)

    -10

  • Option 2)

    10

  • Option 3)

    -25

  • Option 4)

    25

 

Answers (2)

best_answer

As we learned 

 

Double Angle Formula -

Double angle formula

- wherein

These are formulae for double angles.

 

 

 

Addition Formulae -

\tan \left ( A+B \right )= \frac{\tan A+\tan B}{1-\tan A\tan B}

- wherein

A and B are two angles.

 

 

\tan \left ( A+B \right )=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{\frac{10}{3}}{\frac{28}{3}}=\frac{5}{14}

\cos \left ( A+B \right )=2\cos ^{2}\left ( A+B \right )-1\Rightarrow \cos ^{2}\left ( A+B \right )=\frac{196}{221}

given equation =

\cos ^{2}\left ( A+B \right )\left ( 3\tan ^{2} \left ( A+B \right )-10\tan \left ( A+B \right )-25\right )

=-\frac{5525}{196}\times \frac{196}{221}=-25

 

 


Option 1)

-10

Option 2)

10

Option 3)

-25

Option 4)

25

Posted by

Himanshu

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