The refractive index of a prism with apex angle `A` is `cot A//2`. Prove that the angle of minimum deviation is `delta_m = (180^@ - 2 A)`.

Explanation

`mu = (sin (A + delta_(m))//2)/(sin A//2) = "cot"(A)/(2) = (cos A//2)/(sin A//2)`
`:. sin (A + delta_(m))/(2) = cos A//2 = sin (90^(@) - A//2)`
`:. (A + delta_(m))/(2) = 90^(@) - (A)/(2)`
or `(delta_(m))/(2) = 90^(@) - A//2 - A//2 = 90^(@) - A`
or `delta_(m) = 180^(@) - 2 A`