If the refractive index of the material of a prism is $$\cot\left(\frac{A}{2}\right)$$, where $$A$$ is the angle of prism then the angle of minimum deviation will be

$$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$

$$\mu = \cot\left(\frac{A}{2}\right) = \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$

$$\frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \implies \cos\left(\frac{A}{2}\right) = \sin\left(\frac{A + \delta_m}{2}\right)$$

$$\sin\left(\frac{\pi}{2} - \frac{A}{2}\right) = \sin\left(\frac{A + \delta_m}{2}\right) \implies \frac{\pi}{2} - \frac{A}{2} = \frac{A + \delta_m}{2}$$

$$\pi - A = A + \delta_m \implies \delta_m = \pi - 2A$$