Proving $ 4\operatorname{arccot}(2)+\arctan\left(\frac{24}7\right)=\pi $. What am I doing wrong? $$
4\operatorname{arccot}(2)+\arctan\left(\frac{24}7\right)=\pi
$$
original image
To prove the above result, I tried to equate the original expression to some constant $a$ such that $0<a<2.5\pi$ (from the range of the inverse tangent). When I try to solve for $a$ by taking the tangent or sine of both sides, I arrive at the equations:
$$\begin{align}
\sin(a) &=0 \\
\tan(a) &=0
\end{align}$$
which gives me two solutions ($\pi$ and $2\pi$) within the specified range. 
I have already seen other solutions using complex numbers, so I would really appreciate if someone could point out where I'm going wrong rather than a solution via another method.
 A: You need to pay attention to where your angle is. Note that
$$
\operatorname{arccot}(2)=\arctan\left(\frac12\right)\tag1
$$
and that $\arctan\left(\frac12\right)\in\left(0,\frac\pi4\right)$. The identity $\tan(2\arctan(x))=\frac{2x}{1-x^2}$ says
$$
\tan\left(2\arctan\left(\frac12\right)\right)=\frac43\tag2
$$ 
and $2\arctan\left(\frac12\right)\in\left(\frac\pi4,\frac\pi2\right)$ so
$$
2\arctan\left(\frac12\right)=\arctan\left(\frac43\right)\tag3
$$
Thus, $4\arctan\left(\frac12\right)\in\left(\frac\pi2,\pi\right)$ and 
$$
\begin{align}
\tan\left(4\arctan\left(\frac12\right)\right)
&=\tan\left(2\arctan\left(\frac43\right)\right)\tag4\\
&=-\frac{24}7\tag5
\end{align}
$$
Therefore,
$$
4\arctan\left(\frac12\right)=\pi-\arctan\left(\frac{24}7\right)\tag6
$$
Putting together $(1)$ and $(6)$ gives
$$
4\operatorname{arccot}(2)+\arctan\left(\frac{24}7\right)=\pi\tag7
$$
A: Hint:
Use  Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?
and find $$(2+i)^4(7+24i)=\cdots-7^2-24^2$$
Find arguments in both sides
