What is the exact value of #sec^-1 (sqrt2) and csc^-1 (2)#?

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Answer 1 · 2018-03-28T10:39:28

#sec^-1(sqrt2)=pi/4#
#csc^-1(2)=pi/6#

Note:
#color(red)((1)sec^-1x=cos^-1(1/x)#

#color(red)((2)csc^-1x=sin^-1(1/x)#

#color(red)((3)cos^-1(costheta))=color(red)(theta,where,theta in [0,pi]#

#color(red)((4)sin^-1(sintheta))=color(red)(theta,where,theta in [-pi/2,pi/2]#

Here,

#sec^-1(sqrt2)=cos^-1(1/sqrt2)...toApply (1)#

#sec^-1(sqrt2)=cos^-1(cos(pi/4))#

#sec^-1(sqrt2)=pi/4in[0,pi].............toApply(3)#

Now,

#csc^-1(2)=sin^-1(1/2).....toApply(2)#

#csc^-1(2)=sin^-1(sin(pi/6))#

#csc^-1(2)=pi/6in[-pi/2,pi/2]..........toApply(4)#