arctan equation question This question has two parts, I've done the first but I don't understand that second.
a. Show that $arctan(\frac{1}{2})+arctan(\frac{1}{3})=\frac{\pi}{4}$
b. Hence, or otherwise, find the value of $arctan(2)+arctan(3)$.
The mark scheme has a few methods to solve this, and I don't understand this one in particular.
$**arctan(2)+arctan(3)=\frac{\pi}{2}-arctan(\frac{1}{2})+\frac{\pi}{2}-arctan(\frac{1}{3})**$
$=\pi-(arctan(\frac{1}{2})+arctan(\frac{1}{3}))$
$=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$
I don't understand the line that I asterisked. Where does the $\frac{\pi}{2}$ and $arctan(\frac{1}{2})+arctan(\frac{1}{3})$ come from? Please could someone explain all of this to me?
 A: We have for $x>0$:
$$\arctan x+\arctan\left(\frac1x\right)=\frac\pi2$$
and to prove it: let 
$$f(x)=\arctan x+\arctan\left(\frac1x\right)$$
and by differentiate it we find $f'(x)=0$ so $f(x)=f(1)=\frac\pi2,\quad\forall x>0.$ 
A: hints:
Proof that
$$\arctan x=\frac\pi2+\arctan\frac1x\;,\;\;0<x\neq\begin{cases}\frac\pi2\\{}\\\frac2\pi\end{cases}+n\pi\;,\;\;n\in\Bbb Z\;\;:\;$$
$$f(x):=\arctan x+\arctan\frac1x\implies f'(x)=\frac1{1+x^2}-\frac1{x^2}\frac1{1+\frac1{x^2}}=0$$
and thus $\;f(x)=k=$ a constant. Now choose wisely some $\;x\;$ to find out what that constant is...
A: As  $\displaystyle\tan\left(\frac\pi2-z\right)=\cot z,\arctan u+\text{arccot}u=\frac\pi2$
and use Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function? or this, to show that $\displaystyle\arctan x=\text{arccot}\frac1x$ for $x>0$
See also : Proof of $\arctan{2} = \pi/2 -\arctan{1/2}$
A: $$arctg(a)+arcctg(a)=\pi/2$$
and
$$tg(\alpha)=1/ctg(\alpha)$$
A: Sometimes drawing a diagram is helpful.

From the diagram it is clear that
$$\tan A = x$$
$$A = \arctan(x)$$
and that also
$$\tan(\pi/2 - A) = 1/x$$
$$\arctan(\tan(\pi/2 - A)) = \arctan(1/x)$$
$$\pi/2 - A = \arctan(1/x)$$
$$\pi/2 - \arctan(x) = \arctan(1/x)$$
or alternatively,
$$\pi/2 - \arctan(1/x) = \arctan(x)$$
