What does #sin(arccot(5))-sin(arcsec(2))# equal?

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Answer 1 · 2016-07-20T04:22:28

#1/sqrt 26-sqrt 3/2=-0.66991#, nearly.

Use #sin (arc sin( c)) = c#.

Confining to principal values of inverse functions thai are in the 1st

quadrant,

#arc cot(5)=arc tan(1/5)=arc sin (1/sqrt 26# and

#arc sec(2)=arc cos(1/2)=arc sin (sqrt 3/2)#

Now, the given expression is

#sinarc sin (1/sqrt 26)-sin arc sin (sqrt 3/2)#

#=1/sqrt 26- sqrt 3/2=0.6991#, nearly.