Why are the two graph $f(x)=tan(arccos(x/2))$ and $g(x)=sqrt(4-x^2)/x$ equal?

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Answer 1 · 2017-08-24T06:11:04

Please see below.

We have $f(x)=tan(arccos(x/2))$ and let $arccos(x/2)=u$

then $cosu=x/2$ which leads to

$tanu=sqrt(tan^2u)=sqrt(sec^2u-1)=sqrt(1/cos^2u-1)$

= $sqrt(1/(x^2/4)-1)=sqrt(4/x^2-1)=sqrt((4-x^2)/x^2)=sqrt(4-x^2)/x$

Hence $f(x)=tan(arccos(x/2))=sqrt(4-x^2)/x$

Hence $f(x)=tan(arccos(x/2))$ and $g(x)=sqrt(4-x^2)/x$ are same graph.

Why are the two graph f(x)=tan(arccos(x/2))f(x)=tan(arccos(x/2)) and g(x)=sqrt(4-x^2)/xg(x)=sqrt(4-x^2)/x equal?