Simplify $\arctan (\frac{1}{2}\tan (2A)) + \arctan (\cot (A)) + \arctan (\cot ^{3}(A)) $ How to simplify $$\arctan \left(\frac{1}{2}\tan (2A)\right) +  \arctan (\cot (A)) + \arctan (\cot ^{3}(A)) $$ for  $0< A< \pi /4$?
This is one of the problems in a book I'm using. It is actually an objective question , with 4 options given , so i just put $A=\pi /4$ (even though technically its disallowed as $0< A< \pi /4$) and got the answer as $\pi $ which was one of the options , so that must be the answer (and it is weirdly written in options as $4 \arctan (1) $ ).
Still , I'm not able to actually solve this problem. I know the formula for sum of three arctans , but it gets just too messy and looks hard to simplify and it is not obvious that the answer will be constant for all $0< A< \pi /4$. And I don't know of any other way to approach such problems.
 A: Over the given interval we have $\arctan\cot A=\frac{\pi}{2}-A$ and, by setting $t=\tan A$:
$$\begin{eqnarray*}&&\tan\left(\arctan\cot^3 A+\arctan\left(\frac{\tan(2A)}{2}\right)\right)=\frac{\cot^3 A+\frac{1}{2}\tan(2A)}{1-\frac{1}{2}\cot^3 A\tan(2A)}\\&=&\frac{\cot^3 A+\frac{1}{2}\tan(2A)}{1-\frac{1}{2}\cot^3 A\tan(2A)}=\frac{\frac{1}{t^3}+\frac{t}{1-t^2}}{1-\frac{1}{t^2(1-t^2)}}=\frac{1-t^2+t^4}{t(t^2-t^4-1)}=-\frac{1}{t}\end{eqnarray*}$$
so:
$$ \arctan\cot^3 A+\arctan\left(\frac{\tan(2A)}{2}\right)=\frac{\pi}{2}+A $$
and the sum of the three arctangents is $\color{red}{\pi}$ as wanted. Another chance is given by differentiating such a sum wrt to $A$ and check that the derivative is zero oven the given interval, so the sum equals its value in the point $A=\frac{\pi}{8}$, for instance.
A: As $0<A<\dfrac\pi4\implies\cot A>1\implies\cot^3A>1$
Like showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$,
$\arctan(\cot A)+\arctan(\cot^3A)=\pi+\arctan\left(\dfrac{\cot A+\cot^3A}{1-\cot A\cdot\cot^3A}\right)$
Now $\dfrac{\cot A+\cot^3A}{1-\cot A\cdot\cot^3A}=\dfrac{\tan^3A+\tan A}{\tan^4A-1}=\dfrac{\tan A}{\tan^2A-1}=-\dfrac{\tan2A}2$
and $\arctan(-x)=-\arctan(x)$
