Explanation

(c)
Given, $y = \frac{\sin(x + a)}{\sin(x + b)}$
$\Rightarrow \ \ \ \frac{dy}{dx} = \frac{\sin(x + b)\cos{(x + a) - sin(x + b)cos(x + a)}}{\sin^{2}{(a + b)}}$
$\Rightarrow \ \ \ \frac{dy}{dx} = \frac{\sin(b - a)}{\sin^{2}(x + b)}$
For maxima or minima put, $\frac{dy}{dx} = 0$
Sin $(b - a) = 0$
It means no value of$x$, it is neither maxima nor minima.