Line joining the orthocenter to the circumcenter of a triangle ABC is inclined to BC at an angle $\tan^{-1}(\frac{3-\tan B\tan C}{\tan B-\tan C})$ Show that the line joining the orthocenter to the circumscribed center of a triangle ABC is inclined to BC at an angle $\tan^{-1}\left(\frac{3-\tan B\tan C}{\tan B-\tan C}\right)$
I let the foot of perpendicular from A,B,C to opposite sides is D,E,F.Then
$$\tan B=\frac{AD}{BD},\tan C=\frac{AD}{CD}$$
$$\frac{3-\tan B\tan C}{\tan B-\tan C}=\frac{3-\frac{AD}{BD}\frac{AD}{CD}}{\frac{AD}{BD}-\frac{AD}{CD}}$$
I think this way i cannot get answer.Please help me getting the desired proof.I am stuck ...

Let $a$ be the length of side BC coinciding with the x-axis such that vertex B is at origin $(0, 0)$ & vertex C is at $(a, 0)$ then vertex A will at $\left(\frac{a\tan C}{\tan B+\tan C}, \frac{a\tan B\tan C}{\tan B+\tan C}\right)$ Hence, the ortho-center will be $$H\equiv \left(\frac{0+a+\frac{a\tan C}{\tan B+\tan C}}{3}, \frac{0+0+\frac{a\tan B\tan C}{\tan B+\tan C}}{3} \right)\equiv \left(\frac{a(\tan B+2\tan C)}{3(\tan B+\tan C)}, \frac{a\tan B\tan C}{3(\tan B+\tan C)} \right)$$ & the circumscribed center say $D$ can be calculated as $$D\equiv\left(\frac{a}{2}, \frac{a(\tan B\tan C-1)}{2(\tan B+\tan C)} \right)$$
Hence, the slope of the line HD joining $H$ & $D$ with the BC (x-axis) is given as $$m=\frac{y_2-y_1}{x_2-x_1}$$ $$=\frac{\frac{a(\tan B\tan C-1)}{2(\tan B+\tan C)}-\frac{a\tan B\tan C}{3(\tan B+\tan C)}}{\frac{a}{2}-\frac{a(\tan B+2\tan C)}{3(\tan B+\tan C)}}$$ $$=\frac{3\tan B\tan C-3-2\tan B\tan C}{3\tan B+3\tan C-2\tan B-4\tan C}$$ $$=\frac{\tan B\tan C-3}{\tan B-\tan C}$$ Hence the angle of the line joining H & D with side BC is given as $$\tan \theta=|m|$$ $$\implies \theta=\tan^{-1}\left|\frac{\tan B\tan C-3}{\tan B-\tan C}\right|$$
or $$\bbox [5px, border:2px solid #C0A000]{\color{red}{\theta=\tan^{-1}\left|\frac{3-\tan B\tan C}{\tan B-\tan C}\right|}}$$
