Question

If $tan{\theta }_1,tan{\theta }_2,tan{\theta }_3$ are the real roots of $x^3-(a+1)x^2+(b-a)x-b=0where{\theta }_1,{\theta }_2,{\theta }_3$ are acute then ${\theta }_1+{\theta }_2+{\theta }_3=$

• A. $\frac{\pi }{2}$
• B. $\frac{5\pi }{6}$
• C. $\frac{6\pi }{5}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is D

$\frac{\pi }{4}$

$tan{\theta }_1+tan{\theta }_2+tan{\theta }_3=a+1,tan{\theta }_{1}tan{\theta }_{2}tan{\theta }_3=btan{\theta }_{1}tan{\theta }_2+tan{\theta }_{2}tan{\theta }_3+tan{\theta }_{3}tan{\theta }_1=b-a$
$Nowtan({\theta }_1+{\theta }_2+{\theta }_3)=\frac{tan{\theta }_1+tan{\theta }_2+tan{\theta }_3-tan{\theta }_{1}tan{\theta }_{2}tan{\theta }_3}{1-tan{\theta }_{1}tan{\theta }_2-tan{\theta }_{2}tan{\theta }_3-tan{\theta }_{3}tan{\theta }_1}=\frac{a+1-b}{1-(b-a)}=\frac{a-b+1}{1-b+a}=1\therefore {\theta }_1+{\theta }_2+{\theta }_3=\frac{\pi }{4}$