<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] From the given
\begin{align}0=&\frac{1}{2\sqrt 2}- \sin (A-\pi /4) \sin (B-\pi/4) \sin (C-\pi/4)\\
=&\frac{1}{2\sqrt 2}-\frac12\left[ \cos(A-B)-\cos(A+B-\frac\pi2)\right] \sin (C-\pi/4)\\
=&\frac{1}{2\sqrt 2}[ 1- (\cos(A-B)-\sin C) (\sin C - \cos C)]
\end{align}
Note that $\cos(A-B)\le 1$ and
\begin{align}
0 \ge & \frac{1}{2\sqrt 2}[1-( 1 -\sin C ) (\sin C - \cos C)]\\
=& \frac{1}{2\sqrt 2}[(1+\cos C) - \sin C (1+\cos C - \sin C) ]\\
=& \frac{1}{2\sqrt 2}[2\cos^2 \frac C2- \sin C (2\cos^2 \frac C2 - 2\sin \frac C2\cos\frac C2)] \\
=&\frac{1}{\sqrt 2}\cos^2 \frac C2\left[1- 2\sin\frac C2(\cos \frac C2 - \sin \frac C2)\right] \\
=& \frac{1}{\sqrt 2}\cos^2 \frac C2(2- \sin C  - \cos C ) 
\end{align}
which leads to
$\cos\frac C2=0$, i.e.
$C = \pi$ and $A=B = 0$. Thus,
$$
\tan A\tan B+\tan B\tan C+\tan C\tan A = \tan A+\tan B+\tan C =0
$$
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