What does #-2cos(arctan(6))+csc(arcsec(2))# equal?

1 Answer
George C.
Dec 18, 2015

#-2cos(arctan(6))+csc("arcsec"(2))#

#=-2/sqrt(37)+2/sqrt(3)#

#=(-6sqrt(37)+74sqrt(3))/111#

Explanation:

#arctan(6)# is one of the angles in a right angled triangle with legs of length #1#, #6# and hypotenuse #sqrt(1^2+6^2) = sqrt(37)#

Hence #cos(arctan(6)) = 1/sqrt(37)#

#"arcsec"(2)# is one of the angles in a right angled triangle with legs of length #1#, #sqrt(2^2-1^2) = sqrt(3)# and hypotenuse #2#.

Hence #csc("arcsec"(2)) = 2/sqrt(3)#

So:

#-2cos(arctan(6))+csc("arcsec"(2))#

#=-2/sqrt(37)+2/sqrt(3)#

#=-(2sqrt(37))/37+(2sqrt(3))/3#

#=(-6sqrt(37)+74sqrt(3))/111#

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