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

1 Answer
maganbhai P.
Mar 26, 2018

#-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(cot^-1(-6))=sec(pi-cot^-1(6))...to#Apply # color(red)((1)#

#=-sec(cot^-1(6)).......to#Apply #color(red)((2)#

#=-sec(tan^-1(1/6)).........to#Apply #color(red)((3)#

#=-sec(cos^-1(1/(sqrt(1+(1/6)^2))))......to#Apply #color(red)((4)#

#=-sec(cos^-1(1/(sqrt(1+1/36))))#

#=-sec(cos^-1(6/(sqrt(37))))#

#=-sec(sec^-1(sqrt(37)/6)).......to #Apply #color(red)((5)#

#=-sqrt(37)/6...........to#Apply #color(red)((6)#

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