<s>[INST] 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.
 [/INST] 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 $\sin A\sin B-\cos A\cos B=-\cos(A+B)=\cos C$.
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