\[ให้ tan(A)=x,tan(B)=y,tan(C)=z,tan(D)=w\]
\[จาก tan(\alpha+\beta)
=\frac{tan\alpha+tan\beta}{1-(tan\alpha)(tan\beta)}\]
\[จะได้ว่า tan(A+B)=\frac{x+y}{1-xy},tan(C+D)=\frac{w+z}{1-wz}\]
\[ได้ว่าtan(A+B+C+D)=\frac{\frac{x+y}{1-xy}+\frac{w+z}{1-wz}}{...}\]
\[ จากA+B+C+D=360องศา ซึ่ง tan(360องศา)=0 \]
\[ ฉะนั้น {\frac{x+y}{1-xy}+\frac{w+z}{1-wz}}=0 \]
\[\therefore(x+y+z+w)=\sum_{Cyc} xyz \]
\[\because(x+y+z+w)=\sum_{Cyc} xyz \]
\[\therefore\frac{x+y+z+w}{\sum_{Cyc} xyz}=1 \]
\[ คูณ xyzw... \frac{x+y+z+w}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}}=xyzw \]
\[\therefore\frac{tan A + tan B + tan C +tan D}{cot A +cot B + cot C +cot D} = tan A \cdot tan B \cdot tan C \cdot tan D \]