$\left(\frac{\mathrm{si}nA}{1-\cos A}\right)-\left(\frac{1-\cos A}{\sin A}\right)=$

$$\begin{gathered} \frac{\sin \mathrm{A}}{1-\cos \mathrm{A}}-\frac{1-\cos \mathrm{A}}{\sin \mathrm{A}} \\ =\frac{\sin (1+\cos \mathrm{A})}{1-{\cos }^2\mathrm{A}}-\frac{1-\cos \mathrm{A}}{\sin \mathrm{A}} \\ =\frac{\sin \mathrm{A}(1+\cos \mathrm{A})}{{\sin }^2\mathrm{A}}-\frac{1-\cos \mathrm{A}}{\sin \mathrm{A}} \\ =\frac{1+\cos \mathrm{A}-1+\cos \mathrm{A}}{\sin \mathrm{A}} \\ =\frac{2\cos \mathrm{A}}{\sin \mathrm{A}}=2\cot \mathrm{A} \end{gathered}$$