If $f(x)=\frac{\sin (x+\alpha)}{\sin (x+\beta)}, \alpha \neq \beta$, then f(x) has

neither maximum nor minimum

We have, $f(x)=\frac{\sin (x+\alpha)}{\sin (x+\beta)}$

Clearly, $f(x)$ is defined for all $x \neq-\beta$

Now,

$f'(x)=\frac{\sin (x+\beta) \cos (x+\alpha)-\cos (x+\beta) \sin (x+\alpha)}{\sin ^2(x+\beta)}$

$\Rightarrow f'(x)=\frac{\sin (\beta-\alpha)}{\sin ^2(x+\beta)} \neq 0$ as $\alpha \neq \beta$

Hence, $f(x)$ has neither a maximum nor a minimum.