<s>[INST] How to calculate the coordinates of a triangle's orthocentre? How does one calculate the coordinates of a triangle's orthocentre?
I was surfing through the net and got this formula:

$$x-\rm{coordinate}= \frac{x_1\tan A+x_2\tan B+x_3\tan C}{\tan A+\tan B+\tan C}$$
$$y-\rm{coordinate}= \frac{y_1\tan A+y_2\tan B+y_3\tan C}{\tan A+\tan B+\tan C} $$

How do you prove this? Is there any method using just the coordinates?
 [/INST] 
Here $A\equiv (x_1,y_1)$,$B\equiv (x_2,y_2)$,$C\equiv (x_3,y_3)$. I'll use the usual notation for $a,b,c,R$ and $A,B,C$.
Using simple trigonometry, $BP=c\cos B$, $PC=b\cos C\implies\dfrac{BP}{PC}=\dfrac{c\cos B}{b\cos C}$ $$\implies P_x= \dfrac{x_2b\cos C+x_3c\cos B}{b\cos C+c\cos B}= \dfrac{x_2b\cos C+x_3c\cos B}{a}$$ 
Using some trig again, $AH=2R\cos A$ and $HP=2R\cos B \cos C\implies \dfrac{AH}{HP}=\dfrac{\cos A}{\cos B\cos C}$ $$\implies H_x= \dfrac{\cos A\left(\dfrac{x_2b\cos C+x_3c\cos B}{a}\right)+x_1\cos B\cos C}{\cos A+ \cos B\cos C}$$
$$H_x=\dfrac{x_1a\cos B\cos C+x_2b\cos C\cos A+x_3c\cos A\cos B}{a\cos A+a\cos B\cos C}$$
$$H_x=\dfrac{x_1a\cos B\cos C+x_2b\cos C\cos A+x_3c\cos A\cos B}{a\cos B\cos C+b\cos C\cos A+c\cos A\cos B}$$
Dividing by $\cos A\cos B\cos C$:
$$H_x=\frac{x_1a\sec A+x_2b\sec B+x_3c\sec C}{a\sec A+b\sec B+c\sec C}$$
Now using Sine Law do the substitution for $a,b,c$ and then cancel the $2R$ to get:
$$H_x=\dfrac{x_1\tan A+x_2\tan B+x_3\tan C}{\tan A+\tan B+\tan C}$$
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