Explanation

(d)
We have, $a = \cos{\alpha + i\sin{\alpha;}}$
$b = \cos{\beta + i\sin\beta}$
and $c = \cos{\gamma + i\sin\gamma}$
Now, $\frac{b}{c} = \frac{\cos{\beta + i\sin\beta}}{\cos{\gamma + i\sin\gamma}} \times \frac{\cos{\gamma - i\sin\gamma}}{\cos{\gamma - i\sin\gamma}}$
$= \cos{\beta\cos{\gamma + \sin{\beta\sin\gamma}}} + \ i\ \lbrack\sin{\beta\cos{\gamma - \sin{\gamma\cos{\beta\rbrack\ }}}}$
$\Rightarrow \ \frac{b}{c} = \cos{(\beta - \gamma) + i\sin{(\beta - \gamma)}}$ …(i)
Similarly, $\frac{c}{a} = \cos{(\gamma - \alpha) + i\sin{(\gamma - \alpha)}}$ …(ii)
and $\frac{a}{b} = \cos{(\alpha - \beta) + i\sin(\alpha - \beta)}$ …(iii)
On adding Eqs. (i), (ii), (iii), we get
$\cos{(\beta - \gamma) + \cos{(\gamma - \alpha) + \cos(\alpha - \beta)}}$
$+ i{\lbrack sin}{(\beta - \gamma)} + \sin{(\gamma - \alpha) + \sin{(\alpha - \beta)\rbrack\ }} = 1$
On equating real part on both sides, we get
$\cos{(\beta - \gamma) + \cos{(\gamma - \alpha) + \cos(\alpha - \beta)}} = 1$