Answer
See explanation.
Work Step by Step
Use Figure 41.6 in the textbook to relate $\theta_L$ to $L_z$ and L.
$$cos \theta_L=\frac{L_z}{L}$$
$$ \theta_L=arccos\frac{L_z}{L}$$
a. The smallest angle is for the state with the largest L and the largest $L_z$.
This is the state with $l=n-1$ and $m_l=l=n-1$.
$L_z=m_l\hbar=(n-1)\hbar$
$L=\sqrt{l(l+1)}\hbar=\sqrt{(n-1)(n)}\hbar$
$$ \theta_{L,min}=arccos\frac{L_z}{L}=arccos\frac{(n-1)\hbar}{\sqrt{(n-1)n}\hbar}$$
$$ \theta_{L,min}=arccos\frac{(n-1)}{\sqrt{(n-1)n}}$$
This was to be shown. As expected from a classical physics model, it approaches zero degrees for large n.
b. The largest angle is for the state with the largest L and the most negative $L_z$.
This is the state with $l=n-1$ and $m_l=-l=-(n-1)$.
$L_z=m_l\hbar=-(n-1)\hbar$
$L=\sqrt{l(l+1)}\hbar=\sqrt{(n-1)(n)}\hbar$
$$ \theta_{L,max}=arccos\frac{L_z}{L}=arccos\frac{-(n-1)\hbar}{\sqrt{(n-1)n}\hbar}$$
$$ \theta_{L,min}=arccos\frac{(1-n)}{\sqrt{(n-1)n}}$$
As expected from a classical physics model, it approaches 180 degrees for large n.
