# 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?

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}

• Isn’t $\sum \tan A = \tan A \tan B \tan C$? You have written $\tan A \tan B \tan C =\sum \tan A \tan B$ – Aditya Mar 17 at 18:35
• @Aditya: Yes, of course, in \eqref{2} $\sum\tan A\tan B=\sum\tan A=\prod\tan A=\frac{2\rho\,r}{\rho^2-(r+2\,R)^2}$. – g.kov Mar 17 at 18:47

From the given,

$$0=\frac{1}{2\sqrt 2}- \sin (A-\pi /4) \sin (B-\pi/4) \sin (C-\pi/4)$$

$$=\frac{1}{2\sqrt 2}-\frac12\left[ \cos(A-B)-\cos(A+B-\frac\pi2)\right] \sin (C-\pi/4)$$ $$\frac{1}{2\sqrt 2}[ 1- (\cos(A-B)-\sin C) (\sin C - \cos C)]$$

Note that $$\cos(A-B)\le 1$$, and $$0 \ge \frac{1}{2\sqrt 2}[1-( 1 -\sin C ) (\sin C - \cos C)]$$ $$= \frac{1}{2\sqrt 2}[(1+\cos C) - \sin C (1+\cos C - \sin C) ]$$ $$= \frac{1}{2\sqrt 2}[2\cos^2 \frac C2- \sin C (2\cos^2 \frac C2 - 2\sin \frac C2\cos\frac C2)]$$ $$= \frac{1}{\sqrt 2}\cos^2 \frac C2\left[1- 2\sin\frac C2(\cos \frac C2 - \sin \frac C2)\right]$$ $$= \frac{1}{\sqrt 2}\cos^2 \frac C2(2- \sin C - \cos C )$$

So, $$0 \ge \cos^2 \frac C2(2- \sin C - \cos C ) \implies \cos\frac C2=0$$

which leads to $$C = \pi$$ and $$A=B = 0$$. Thus,

$$\tan A\tan B+\tan B\tan C+\tan C\tan A = \tan A+\tan B+\tan C =0$$