How do you evaluate #tan(sec^-1 4)#?

1 Answer
Bdub
Oct 13, 2016

#tan(sec^-1 4)=sqrt15#

Explanation:

#tan(sec^-1 4)#

The restriction for the range of #sec^-1x# is #[0,pi],y!=pi/2 # and since the argument is positive it means that our triangle is in quadrant I with the hypotenuse of 4 and adjacent of 1 and therefore using pythagorean theorem we have the opposite is #sqrt15#.

Therefore, the ratio for tangent from our triangle is
#tan theta=o/a=sqrt15/1=sqrt15#

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