If H is orthocenter of the triangle ABC, then AH is equal to:

from $ΔAHB$

$\frac{AH}{\sin(90-A)}=\frac{AB}{\sin(A+B)}=\frac{BH}{sin(90-B)}$

$⇒\frac{AH}{\sin(90-A)}=\frac{AB}{\sin(A+B)}=\frac{C}{sin(180-(A+B))}$

$⇒\frac{AH}{\cos A}=\frac{c}{\sin C}=\frac{a}{\sin A}$

$⇒AH=a\cot A$