We can use the Pythagorean identity to find $\sin\theta$:
$$\sin^2\theta=1-\cos^2\theta=1-\frac{u^2}{u^2+5}=\frac{5}{u^2+5}$$
$$\sin\theta=\pm\sqrt{\frac{5}{u^2+5}}$$
Since $u>0$, we know that $\cos^{-1}\frac{u}{\sqrt{u^2+5}}$ is in the first or fourth quadrant,
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