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Find an algebraic expression for \(tan(arcsin(\frac{\sqrt{36-x^2}}{6}))\).

I don't understand why the answer has to be simplified from \(\frac{\sqrt{36-x^2}}{\sqrt{x^2}}\) to \(\frac{\sqrt{36-x^2}}{x}\) since radical x2 isn't equal to x when x is negative.

 Mar 21, 2019
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The derivation of the answer is based on a right triangle with opposite leg equal to x.

 

Triangles don't have negative length legs. 

 

But you are correct that it should be mentioned that 0<= x <=6

 Mar 21, 2019

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