The angle $ \angle APB$ between the tangents to a curve. 
I have been asked to find $\angle{APB}$ in the form: $\tan^{-1}{\alpha}$ + $\tan^{-1}{\beta}$
I got the equations of the lines through differentiating the function $h(x) = {(\ln(x) - 1.5)}^{2} - 0.25$ at $e$ and ${e}^{2}$ respectively
The equation of the first line is: $$y = -xe^{-1} + 1$$
The equation of the second line is: $$y = xe^{-2} -1$$
Then after doing some math, I got:
$\angle{APB}$ = 180 - (180 - $\tan^{-1}{(\frac{-1}{e})}$ + $\tan^{-1}{(\frac{1}{e^2})}$)
$\angle{APB}$ = $\tan^{-1}{(\frac{-1}{e})}$ - $\tan^{-1}{(\frac{1}{e^2})}$
However, this is not the correct answer(due to the minus sign). I do not know if I am doing the correct work so far or if I have gone far off.
$ \angle APB$

In any triangle external angle made by producing a line of  any triangle equals the sum of two opposite angles.
$$ \gamma_2- \gamma_1 $$
made at x-coordinate locations $(e^2,e) $ respectively
As you marked, adopting a consist anti clockwise rotation convention reckoned positive
$$ \pi+\tan^{-1} \alpha  - (\pi-\tan^{-1} \beta )$$
$$ \tan^{-1} \alpha  + \tan^{-1} \beta $$
Now $ \beta <0,$ then only can you get arctan obtuse between $(\pi/2,\pi)  $ in second quadrant.
Numerical calculation results in a value  $\approx 152^{\approx}.$
