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

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Answer 1 · 2015-12-18T22:37:49

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

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

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

$arctan(6)$ 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:

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

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

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

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

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