A) 25
B) -25
C) -10
D) 10
Correct Answer: B
Solution :
\[\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\] Using the fact that \[\tan A\] and \[\tan B\] are the roots of \[3{{x}^{2}}-10x-25=0\], we get \[\tan (A+B)=\frac{10/3}{28/3}=\frac{5}{14}\] We also see that \[\begin{align} & \cos 2(A+B)=-1+2{{\cos }^{2}}(A+B)= \\ & \frac{1-{{\tan }^{2}}(A+B)}{1+{{\tan }^{2}}(A+B)}\Rightarrow {{\cos }^{2}}(A+B)=\frac{196}{221} \\ \end{align}\] We see that \[\begin{align} & 3{{\sin }^{2}}(A+B)-10\sin (A+B)\cos (A+B)-25{{\cos }^{2}}(A+B)= \\ & {{\cos }^{2}}(A+B)(3ta{{n}^{2}}(A+B)-10\tan (A+B)-25) \\ \end{align}\] \[=\frac{75-700-4900}{196}\times \frac{121}{221}=-\frac{5525}{196}\times \frac{196}{221}=-25\] So option B is the correct answer.
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