From the given figure, find the value of:
$\tan A-\cot C$

First we have to find the value of $BC$
So, we can find the value of $BC$ with the help of Pythagoras theorem
According to Pythagoras theorem
${\left(Hypotenuse\right)}^2={\left(Base\right)}^2+{\left(Perpendicular\right)}^2$
$\Rightarrow {\left(AB\right)}^2+{\left(BC\right)}^2={\left(AC\right)}^2\Rightarrow {\left(12\right)}^2+{\left(BC\right)}^2={\left(13\right)}^2\Rightarrow 144+{\left(BC\right)}^2=169\Rightarrow {\left(BC\right)}^2=169-144=25$
$\Rightarrow {\left(BC\right)}^2=25\Rightarrow BC=\sqrt{25}=\pm 5$
But side $BC$ can't be negative so, $BC=5$