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