0
So do this by a right triangle.
tan θ=tan arcsecx
θ=arcsecx
x=secθ
x=1/cosθ
cosθ=1/x
In a right triangle, this would look like the following
___________________________________e
________________________________s__o
____________________________u______p
_____hypotenuse=1________n_________p____opposite=[1-(1/x^2)]^(1/2)
____________________e______________o
________________t__________________s
____________o______________________i
________p__________________________t
____y__θ________________________e
h____a____d____j____a____c___e___n__t
adjacent=1/x
the tangent of theta is then the length of the opposite over the adjacent, so it is
{[1-(1/x^2)]^(1/2)}/(1/x)
This is going to simplify to
[(x^2)-1]^(1/2)
That's how you do it. Sorry about all the parentheses and lines in the triangle, but I don't know how to do it any other way.
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∙ 14y ago
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