Question

If $\tan A,\tan B,\tan C$ are the roots of the equation $x^3-3x^2-2x+1=0,$ then the value of ${\sin }^2(A+B+C)+\sin (A+B+C)\cos (A+B+C)$ is

• A. $0$
• B. $\frac{4}{5}$
• C. $\frac{13}{21}$
• D. $\frac{28}{25}$

Answer 1

The correct option is D $\frac{28}{25}$

Let $A+B+C=\alpha$ and
$S={\sin }^2\alpha +\sin \alpha \cos \alpha ={\cos }^2\alpha [{\tan }^2\alpha +\tan \alpha ]$

Now, from the given equation
$x^3-3x^2-2x+1=0$
Sum of the roots
$\sum \tan A=3$
Sum of roots taken two at a time,
$\sum \tan A\tan B=-2$
Product of the roots,
$\prod \tan A=-1$

Now,
$\tan (A+B+C)=\frac{\sum \tan A-\prod \tan A}{1-\sum \tan A\tan B}=\frac{3-(-1)}{1-(-2)}=\frac{4}{3}$
Therefore,
$\tan \alpha =\frac{4}{3}\Rightarrow {\cos }^2\alpha ={\left(\frac{3}{5}\right)}^2=\frac{9}{25}$

Now,
$S={\cos }^2\alpha [{\tan }^2\alpha +\tan \alpha ]=\frac{9}{25}[\frac{16}{9}+\frac{4}{3}]$

$\therefore S=\frac{28}{25}$