Angle is in first quadrant and we know that in first quadrant all six trigonometric functions are positive.

Given, \(\displaystyle{\cos{{u}}}=\frac{{5}}{{13}}\)

We have identity \(\displaystyle{{\sin}^{{2}}{x}}+{{\cos}^{{2}}{x}}={1}\)

Therefore,

\(\displaystyle{\sin{{u}}}=\sqrt{{{1}-{{\cos}^{{2}}{u}}}}\)

\(\displaystyle=\sqrt{{{1}-{\left(\frac{{5}}{{13}}\right)}^{{2}}}}\)

\(\displaystyle=\sqrt{{{1}-\frac{{25}}{{169}}}}\)

\(\displaystyle=\sqrt{{\frac{{{169}-{25}}}{{169}}}}\)

\(\displaystyle=\sqrt{{\frac{{144}}{{169}}}}\)

\(\displaystyle=\frac{{12}}{{13}}\)

Now, finding other trigonometric function.

\(\displaystyle{\tan{{u}}}=\frac{{{\sin{{u}}}}}{{{\cos{{u}}}}}\)

\(\displaystyle=\frac{{\frac{{12}}{{13}}}}{{\frac{{5}}{{13}}}}\)

\(\displaystyle=\frac{{12}}{{5}}\)

\(\displaystyle{\sec{{u}}}=\frac{{1}}{{{\cos{{u}}}}}\)

\(\displaystyle=\frac{{1}}{{\frac{{5}}{{13}}}}\)

\(\displaystyle=\frac{{13}}{{5}}\)

\(\displaystyle{\cos{{e}}}{c}{u}=\frac{{1}}{{{\sin{{u}}}}}\)

\(\displaystyle=\frac{{1}}{{\frac{{12}}{{13}}}}\)

\(\displaystyle=\frac{{13}}{{12}}\)

\(\displaystyle{\cot{{u}}}=\frac{{1}}{{{\tan{{u}}}}}\)

\(\displaystyle=\frac{{1}}{{\frac{{12}}{{5}}}}\)

\(\displaystyle=\frac{{5}}{{12}}\)

Given, \(\displaystyle{\cos{{u}}}=\frac{{5}}{{13}}\)

We have identity \(\displaystyle{{\sin}^{{2}}{x}}+{{\cos}^{{2}}{x}}={1}\)

Therefore,

\(\displaystyle{\sin{{u}}}=\sqrt{{{1}-{{\cos}^{{2}}{u}}}}\)

\(\displaystyle=\sqrt{{{1}-{\left(\frac{{5}}{{13}}\right)}^{{2}}}}\)

\(\displaystyle=\sqrt{{{1}-\frac{{25}}{{169}}}}\)

\(\displaystyle=\sqrt{{\frac{{{169}-{25}}}{{169}}}}\)

\(\displaystyle=\sqrt{{\frac{{144}}{{169}}}}\)

\(\displaystyle=\frac{{12}}{{13}}\)

Now, finding other trigonometric function.

\(\displaystyle{\tan{{u}}}=\frac{{{\sin{{u}}}}}{{{\cos{{u}}}}}\)

\(\displaystyle=\frac{{\frac{{12}}{{13}}}}{{\frac{{5}}{{13}}}}\)

\(\displaystyle=\frac{{12}}{{5}}\)

\(\displaystyle{\sec{{u}}}=\frac{{1}}{{{\cos{{u}}}}}\)

\(\displaystyle=\frac{{1}}{{\frac{{5}}{{13}}}}\)

\(\displaystyle=\frac{{13}}{{5}}\)

\(\displaystyle{\cos{{e}}}{c}{u}=\frac{{1}}{{{\sin{{u}}}}}\)

\(\displaystyle=\frac{{1}}{{\frac{{12}}{{13}}}}\)

\(\displaystyle=\frac{{13}}{{12}}\)

\(\displaystyle{\cot{{u}}}=\frac{{1}}{{{\tan{{u}}}}}\)

\(\displaystyle=\frac{{1}}{{\frac{{12}}{{5}}}}\)

\(\displaystyle=\frac{{5}}{{12}}\)