(a) As we know, the refractive index of the material of the prism
$$
\begin{aligned}
&\mu=\frac{\sin \left(\frac{\delta_{m}+A}{2}\right)}{\sin (A / 2)} \\
&\cot A / 2=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin A / 2}=\frac{\cos (A / 2)}{\sin (A / 2)} \\
&\Rightarrow \operatorname{Sin}\left(\frac{\delta_{m}+A}{2}\right)=\sin \left(90^{\circ}+A / 2\right) \\
&\Rightarrow \delta_{\min }=180^{\circ}-2 A
\end{aligned}
$$
