Question

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

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\frac{3\pi }{4}$
• D. $\pi$

Answer 1

The correct option is B $\frac{\pi }{4}$

$\tan {\theta }_1+\tan {\theta }_2+\tan {\theta }_3=(a+1)$

$\Sigma \tan {\theta }_1\tan {\theta }_2=(b-a)$

$\tan {\theta }_1\tan {\theta }_2\tan {\theta }_3=b$

using identity
$\tan ({\theta }_1+{\theta }_2+{\theta }_3)=\frac{\Sigma \tan {\theta }_1-\Pi \tan {\theta }_1}{1-\Sigma \tan {\theta }_1\tan {\theta }_2}$

$=\frac{a+1-b}{1-(b-a)}=1$

$\Rightarrow {\theta }_1+{\theta }_2+{\theta }_3=\frac{\pi }{4}$