Problem source

The positive numbers $a$, $b$ and $c$ satisfy
$bc=a^2+1$. Prove that
$$
\arctan\left(\frac1  {a+b}\right)+
\arctan\left(\frac1 {a+c}\right)=
\arctan\left(\frac1 a   \right).
$$
The positive numbers $p$, $q$, $r$, $s$, $t$, $u$ and $v$ satisfy
$$
st = (p+q)^2 + 1 \;, \ \ \ \ \ \ uv=(p+r)^2 + 1 \;, \ \ \ \ \ \ 
qr = p^2+1\;.
$$
Prove that 
$$
\arctan  \! \!\left(\!\frac1 {p+q+s}\!\right)  + 
\arctan  \! \!\left(\!\frac 1{p+q+t}\!\right)   + 
\arctan  \! \!\left(\!\frac 1 {p+r+u}\!\right)  + 
\arctan  \! \!\left(\!\frac1  {p+r+v}\!\right) 
=\arctan \! \!\left(    \! \frac1 p \! \right) .
$$ 
Hence show that
$$
\arctan\left(\frac1  {13}\right)
+\arctan\left(\frac1 {21}\right)
+\arctan\left(\frac1  {82}\right)
+\arctan\left(\frac1 {187}\right)
=\arctan\left(\frac1  {7}\right).
$$
[Note that $\arctan x$ is another notation for $ \tan^{-1}x \,.\,$]

Solution source

\begin{align*}
&& \tan \left (\arctan\left(\frac1  {a+b}\right)+
\arctan\left(\frac1 {a+c}\right) \right) &= \frac{\frac1{a+b}+\frac1{a+c}}{1-\frac{1}{(a+b)(a+c)}} \\
&&&= \frac{a+c+a+b}{(a+b)(a+c)-1} \\
&&&= \frac{2a+b+c}{a^2+ab+ac+bc-1} \\
&&&= \frac{2a+b+c}{2a^2+ab+ac} \\
&&&= \frac{1}{a} \\
&&&= \tan \arctan \frac1a\\
\Rightarrow && \arctan\left(\frac1  {a+b}\right)+
\arctan\left(\frac1 {a+c}\right) &= \arctan \frac{1}{a} + n \pi
\end{align*}
Since $\arctan x \in (-\frac{\pi}{2}, \frac{\pi}{2})$ the LHS $\in (0, \pi)$ so $n = 0$.

\begin{align*}
a=p+q, b = s, c = t:&& \arctan  \! \!\left(\!\frac1 {p+q+s}\!\right)  + 
\arctan  \! \!\left(\!\frac 1{p+q+t}\!\right)  &= \arctan \left ( \frac{1}{p+q} \right) \\
a=p+r, b= u, c = v && \arctan  \! \!\left(\!\frac 1 {p+r+u}\!\right)  + 
\arctan  \! \!\left(\!\frac1  {p+r+v}\!\right)  &= \arctan  \! \!\left(\!\frac1  {p+r}\!\right) \\
a = p, b = q, c = r:&& \arctan \left ( \frac{1}{p+q} \right) +\arctan  \! \!\left(\!\frac1  {p+r}\!\right) &= \arctan \left ( \frac1p \right)
\end{align*}

and the result follows.

Taking $p = 7$ we need to solve \[ \begin{cases} q+s &= 6 \\ q+t &= 14 \\ r+u &= 75 \\ r+v &= 180 \end{cases} \] also satisfying $qr = 50$ etc, so say $q  = 1, r = 50, s = 5, v=25$