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

I know about some basic conditional identities but don't know how to use them here.
 A: 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 $-k=\cos{\pi}=-1 \Rightarrow k=1$
A: $$\dfrac{\sin A\sin B\cos C+\sin B\sin C\cos A+\sin C\sin A\cos B-1}{\cos A\cos B\cos C}$$
As $C+A=\pi-B,\sin(C+A)=\sin B,\cos(C+A)=-\cos B$
Now,$\sin A\sin B\cos C+\sin B\sin C\cos A$
$=\sin B\sin(C+A)=\sin^2B=1-\cos^2B=1+\cos B\cos(C+A)$
$\sin A\sin B\cos C+\sin B\sin C\cos A+\sin C\sin A\cos B$
$=1+\cos B\cos(C+A)+\sin C\sin A\cos B$
$=1+\cos B\{\cos(C+A)+\sin C\sin A\}$
$=1+\cos B(\cos C\cos A-\sin C\sin A+\sin C\sin A)$
$=1+\cos A\cos B\cos C$
Can you take it from here?
A: 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 $\sin A\sin B-\cos A\cos B=-\cos(A+B)=\cos C$.
