$$ N \cos\theta - \mu N \sin\theta = mg $$

$$ N \sin\theta + \mu N \cos\theta = \frac{mv_0^2}{r} $$

Dividing the second equation by the first gives

$$ \frac{v_0^2}{r} = \frac{g\left(\sin\theta + \mu \cos\theta\right)}{\cos\theta - \mu \sin\theta}. $$

Multiplying both sides by $\cos\theta - \mu\sin\theta$ leads to

$$ \frac{v_0^2}{r}(\cos\theta - \mu \sin\theta) = g(\sin\theta + \mu \cos\theta). $$

Expanding and grouping the terms with $\mu$:

$$ \frac{v_0^2}{r}\cos\theta - g\sin\theta = \mu\left(\frac{v_0^2}{r}\sin\theta + g\cos\theta\right). $$

Thus, solving for $\mu$:

$$ \mu = \frac{\frac{v_0^2}{r}\cos\theta - g\sin\theta}{\frac{v_0^2}{r}\sin\theta + g\cos\theta}. $$

Multiplying the numerator and the denominator by $r$ gives

$$ \mu = \frac{v_0^2\cos\theta - rg\sin\theta}{v_0^2\sin\theta + rg\cos\theta}. $$

Dividing both the numerator and denominator by $\cos\theta$ (assuming $\cos\theta \neq 0$):

$$ \mu = \frac{v_0^2 - rg\tan\theta}{v_0^2\tan\theta + rg}. $$

This result matches the expression

$$ \mu=\frac{v_0^2 - rg\,\tan\theta}{rg + v_0^2\,\tan\theta}. $$

Thus, the correct answer is:

Option C