How do you find the value of sec [Cot^-1 (-6)]?

Mar 26, 2018

$- \frac{\sqrt{37}}{6}$

Explanation:

We know that,
color(red)((1)cot^-1(-x)=pi-cot^-1x
color(red)((2)sec(pi-theta)=-sectheta
color(red)((3)cot^-1x=tan^-1(1/x), x > 0
color(red)((4)tan^-1x=cos^-1(1/(sqrt(1+x^2)))
color(red)((5)cos^-1x=sec^-1(1/x)
color(red)((6)sec(sec^-1x)=x
Here,

$\sec \left({\cot}^{-} 1 \left(- 6\right)\right) = \sec \left(\pi - {\cot}^{-} 1 \left(6\right)\right) \ldots \to$Apply  color(red)((1)

$= - \sec \left({\cot}^{-} 1 \left(6\right)\right) \ldots \ldots . \to$Apply color(red)((2)

$= - \sec \left({\tan}^{-} 1 \left(\frac{1}{6}\right)\right) \ldots \ldots \ldots \to$Apply color(red)((3)

$= - \sec \left({\cos}^{-} 1 \left(\frac{1}{\sqrt{1 + {\left(\frac{1}{6}\right)}^{2}}}\right)\right) \ldots \ldots \to$Apply color(red)((4)

$= - \sec \left({\cos}^{-} 1 \left(\frac{1}{\sqrt{1 + \frac{1}{36}}}\right)\right)$

$= - \sec \left({\cos}^{-} 1 \left(\frac{6}{\sqrt{37}}\right)\right)$

$= - \sec \left({\sec}^{-} 1 \left(\frac{\sqrt{37}}{6}\right)\right) \ldots \ldots . \to$Apply color(red)((5)

$= - \frac{\sqrt{37}}{6.} \ldots \ldots \ldots . \to$Apply color(red)((6)