SOLUTION
Given
[tex]tan\mleft(cos^{-1}\mleft(\frac{u}{3}\mright)\mright)[/tex]
Using the following identity
[tex]tan\mleft(cos^{-1}\mleft(x\mright)\mright)=\frac{\sqrt[]{1-x^2}}{x}[/tex]
It follows:
[tex]tan(cos^{-1}(\frac{u}{3}))=\frac{\sqrt[]{1-(\frac{u}{3})^2}}{\frac{u}{3}}[/tex]
Simplify the right hand side
[tex]\frac{\sqrt[]{1-(\frac{u}{3})^2}}{\frac{u}{3}}=\frac{\sqrt[]{1-\frac{u^2}{9}}}{\frac{u}{3}}[/tex]
This can further be reduce
[tex]\begin{gathered} \frac{\sqrt[]{1-\frac{u^2}{9}}}{\frac{u}{3}} \\ =\frac{\sqrt[]{\frac{9-u^2}{9}}}{\frac{u}{3}} \\ =\frac{\frac{1}{3}\sqrt[]{9-u^2}}{\frac{u}{3}} \\ =\frac{\sqrt[]{9-u^2}}{3\times\frac{u}{3}} \\ =\frac{\sqrt[]{9-u^2}}{u} \end{gathered}[/tex]
Therefore the solution is:
[tex]\frac{\sqrt[]{9-u^2}}{u}[/tex]
