Given \(\cos \theta = \frac{x}{y}\),

We know \(\sin^2 \theta + \cos^2 \theta = 1\),

So \(\sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\frac{x}{y}\right)^2 = \frac{y^2 - x^2}{y^2}\).

Assuming \(\sin \theta > 0\) (acute angle or principal value),

\(\sin \theta\) = \(\frac{\sqrt{y^2 - x^2}}{\text{y}}\).

Then, \(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sqrt{y^2 - x^2}/y}{x/y} = \frac{\sqrt{y^2 - x^2}}{x}\).

Thus, \(\tan \theta = \frac{\sqrt{y^2 - x^2}}{x}\)