Question

If $\alpha$ is one of the roots of $x^2+7x+10=0$, then the value of ${\cot }^{-1}\alpha +{\cot }^{-1}\frac{1}{\alpha }$ is equal to

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

Answer 1

The correct option is C $\frac{3\pi }{2}$

Given: $x^2+7x+10=0$
$\Rightarrow x=-2,-5$
Since, both roots of $x^2+7x+10=0$ is negative.
${\cot }^{-1}\alpha =\pi +{\tan }^{-1}\frac{1}{\alpha };\alpha <0$
$\Rightarrow {\cot }^{-1}\alpha +{\cot }^{-1}\frac{1}{\alpha }$
$={\cot }^{-1}\alpha +\pi +{\tan }^{-1}\alpha$
$=\pi +({\tan }^{-1}\alpha +{\cot }^{-1}\alpha )$
$=\pi +\frac{\pi }{2}=\frac{3\pi }{2}$