How do you write an algebraic expression that is equivalent to #tan(arccos(x/3))#?

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Answer 1 · 2018-06-21T13:58:47

The answer is #=+-sqrt(9-x^2)/x#

Let #y=arccos(x/3)#

Then,

#x/3=cosy#

#tan^2y+1=1/cos^2y#

#tan^2y=1/(x^2/9)-1=9/x^2-1=(9-x^2)/x^2#

#tany=+-sqrt((9-x^2)/x^2)=+-sqrt(9-x^2)/x#

Finally,

#tan(arccos(x/3))=+-sqrt(9-x^2)/x#