Answer
$$0$$
Work Step by Step
Recall the co-function identity:
$ \sin (\theta)=\cos(90^{\circ}-\theta)$ and $\cot \theta=\dfrac{\cos \theta}{\sin \theta}$
Therefore, $$\cot \ 40^{\circ} -\dfrac{\sin 50^{\circ}}{\sin 40^{\circ}}=\cot \ 40^{\circ} -\dfrac{\cos 40^{\circ}}{\sin 40^{\circ}}\\=\cot \ 40^{\circ} -\cot 40^{\circ}\\=0$$
