Question

If $\tan A$ and $\tan B$ are the roots of the quadratic equation, $3x^2-10x-25=0$, then the value of $3{\sin }^2(A+B)-10\sin (A+B).\cos (A+B)-25{\cos }^2(A+B)$

• A. $-10$
• B. $10$
• C. $-25$
• D. $25$

Answer 1

Answer: (C)

$-25$

Using the fact that $\tan A$ and $\tan B$ are the roots of $3x^2-10x-25=0$,
Sum of the roots $=\frac{10}{3}$ Product of the roots $=\frac{-25}{3}$
$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$
we get , $\tan (A+B)=\frac{10/3}{28/3}=\frac{5}{14}$

We have,

$\cos 2(A+B)=-1+2{\cos }^2(A+B)=\frac{1-{\tan }^2(A+B)}{1+{\tan }^2(A+B)}=\frac{1-\frac{25}{196}}{1+\frac{25}{196}}\Rightarrow {\cos }^2(A+B)=\frac{196}{221}$

We see that

$3{\sin }^2(A+B)-10\sin (A+B)\cos (A+B)-25{\cos }^2(A+B)$

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

$=\frac{75-700-4900}{196}\times \frac{121}{221}$

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

So option C is the correct answer.