<s>[INST] If $A, B, C$ are angles of $\Delta ABC$ and $\sin (A-\pi /4) \sin (B-\pi/4) \sin (C-\pi/4)=\frac{1}{2\sqrt 2}$... 
If $A, B, C$ are angles of $\Delta ABC$ and $\sin (A-\pi /4) \sin (B-\pi/4) \sin (C-\pi/4)=\frac{1}{2\sqrt 2}$, then prove that $\sum \tan A \tan B=\sum \tan A$

Solving the given equation, we get$$(\sin A-\cos A)(\sin B -\cos B)(\sin C -\cos C)=1$$
$$(\tan A-1)(\tan B-1)(\tan C-1)=\sec A \sec B \sec C$$
$$\sum \tan A -\sum \tan A \tan B +\sum \tan A-1=\sec A \sec B \sec C$$
How should I proceed?
 [/INST] Hint:
Both
\begin{align} 
\sin (A-\pi /4) \sin (B-\pi/4) \sin (C-\pi/4)=\frac{1}{2\sqrt 2}
\tag{1}\label{1}
\end{align} 
and
\begin{align} 
\tan A\tan B+\tan B\tan C+\tan C\tan A
&=
\tan A+\tan B+\tan C
,
\tag{2}\label{2}
\end{align}
expressed in terms of semiperimeter $\rho$, 
inradius $r$ and circumradius $R$ of the triangle,
are equivalent to
\begin{align} 
\rho^2&=r^2+4\,r\,R+2\,\rho\,r
\tag{3}\label{3}
,
\end{align} 
which implies
\begin{align} 
\rho&=r+\sqrt{2\,r^2+4\,r\,R}
\tag{4}\label{4}
.
\end{align} 
But \eqref{4} is true only for the degenerate triangle
when one angle is $180^\circ$ and the other two are zero.
For the conversion, 
use known identities for the angles of triangle
\begin{align}
\cos A\cos B\cos C&=\frac{r}{R}+1
\tag{5}\label{5}
,\\
\sin A\sin B\sin C
&=
\frac{\rho\,r}{2R^2}
\tag{6}\label{6}
,
\end{align}
\begin{align}
\tan A\tan B\tan C
=
\tan A+\tan B+\tan C
&=\frac{2\rho\,r}{\rho^2-(r+2\,R)^2}
\tag{7}\label{7}
,\\
\tan A\tan B+\tan B\tan C+\tan C\tan A
&=1+\frac{4\,R^2}{\rho^2-(r+2\,R)^2}
\tag{8}\label{8}
,\\
\cot A+\cot B+\cot C&=
\frac12\,\left(\frac{\rho}r
-\frac r\rho \right)
-2\,\frac R\rho
\tag{9}\label{9}
.
\end{align}
</s>