Given $A+B+C=180^{\circ}$, find value of $\tan A\cdot\tan B+\tan B\cdot\tan C+\tan A\cdot \tan C-\sec A\cdot\sec B\cdot\sec C$

Observe that

$$\cos{A+B+C}=\cos{A}\cos{B}\cos{C}-\sin{A}\sin{B}\cos{C}-\sin{A}\cos{B}\sin{C}-\cos{A}\sin{B}\sin{C}\tag{1}$$

Let $k=\tan A\cdot\tan B+\tan B\cdot\tan C+\tan A\cdot \tan C-\sec A\cdot\sec B\cdot\sec C$

Divide $(1)$ by $\cos^2{A}\cos^2{B}\cos^2{C}$, to get the $-k=\cos{\pi}=-1 \Rightarrow k=1$

If you write $\tan C=\tan(\pi-(A+B))=-\tan(A+B)$, you have \begin{align} \tan A\tan B+\tan B\tan C+\tan C\tan A &= \tan A\tan B-(\tan A+\tan B)\frac{\tan A+\tan B}{1-\tan A\tan B} \\[6px] &= \frac{ \tan A\tan B-\tan^2A\tan^2B-\tan^2A-2\tan A\tan B-\tan^2B }{1-\tan A\tan B} \\[6px] &= \frac{(\tan^2A\tan^2B+\tan^2A+\tan^2B+1)+(\tan A\tan B-1)} {\tan A\tan B-1} \\[6px] &= \frac{1}{\tan A\tan B-1}(\tan^2A+1)(\tan^2B+1)+1 \\[6px] &= \frac{\cos A\cos B}{\sin A\sin B-\cos A\cos B} \frac{1}{\cos^2A} \frac{1}{\cos^2B} +1 \\[6px] &= \frac{1}{\cos A\cos B\cos C}+1 \end{align} because $\sin A\sin B-\cos A\cos B=-\cos(A+B)=\cos C$.