If $\mathrm{A}+\mathrm{B}+\mathrm{C}={360}^{\circ }$ then $\cot \frac{\mathrm{A}}{4}+\cot \frac{\mathrm{B}}{4}+\cot \frac{\mathrm{C}}{4}=$ 

$\mathrm{A}+\mathrm{B}+\mathrm{C}={360}^0$

$\frac{\mathrm{A}}{4}+\frac{\mathrm{B}}{4}+\frac{\mathrm{C}}{4}={90}^{\circ }\Rightarrow \frac{\mathrm{A}}{4}+\frac{\mathrm{B}}{4}={90}^{\circ }-\frac{\mathrm{C}}{4}$

$\Rightarrow \cot \left(\frac{\mathrm{A}}{4}+\frac{\mathrm{B}}{4}\right)=\cot \left(90-\frac{\mathrm{C}}{4}\right)$

$\Rightarrow \frac{\cot \frac{\mathrm{A}}{4}\cot \frac{\mathrm{B}}{4}-1}{\cot \frac{\mathrm{B}}{4}+\cot \frac{\mathrm{A}}{4}}=\frac{1}{\cot \frac{\mathrm{C}}{4}}\Rightarrow \cot \frac{\mathrm{A}}{4}\cot \frac{\mathrm{B}}{4}\cot \frac{\mathrm{C}}{4}-\cot \frac{\mathrm{C}}{4}=\cot \frac{\mathrm{A}}{4}+\cot \frac{\mathrm{B}}{4}$

$\therefore \cot \frac{\mathrm{A}}{4}+\cot \frac{\mathrm{B}}{4}+\cot \frac{\mathrm{C}}{4}=\cot \frac{\mathrm{A}}{4}\cot \frac{\mathrm{B}}{4}\cot \frac{\mathrm{C}}{4}$