Question

If $x_{1}, x_{2}, x_{3}, x_{4}$ are roots of the equation $x^{4}-x^{3} \sin 2 \beta+$ $x^{2} \cos 2 \beta-x \cos \beta-\sin \beta=0$, then $\displaystyle\sum_{i=1}^{4} \tan ^{-1} x_{i}$ is equal to

• A. $\beta$
• B. $\frac{\pi}{2}-\beta$
• C. $\pi-\beta$
• D. $-\beta$

Answer 1

Answer: (B)

$\Sigma x _{1}=\sin 2 \beta, \Sigma x _{1} x _{2}$
$=\cos 2 \beta, \Sigma x _{1}\, x _{2} \,x _{3}=\cos \beta$
and $ x_{1} \,x_{2} \,x_{3} \,x_{4}=-\sin \beta $
$ \therefore \tan ^{-1} x_{1}+\tan ^{-1} x_{2}+\tan ^{-1} x_{3}+\tan ^{-1} x_{4}$
$=\tan ^{-1}\left(\frac{\sum x _{1}-\sum x _{1} \,x _{2} \,x _{3}}{1-\sum x _{1}\, x _{2}+ x _{1}\, x _{2} \,x _{3} x _{4}}\right)$
$=\tan ^{-1}\left(\frac{\sin 2 \beta-\cos \beta}{1-\cos 2 \beta-\sin \beta}\right)$
$=\tan ^{-1}\left\{\frac{\cos \beta(2 \sin \beta-1)}{\sin \beta(2 \sin \beta-1)}\right\}=\tan ^{-1} \cot \beta$
$=\tan ^{-1} \tan \left(\frac{\pi}{2}-\beta\right)$
$=\frac{\pi}{2}-\beta$