The positive numbers \(a\), \(b\) and \(c\) satisfy \(bc=a^2+1\). Prove that \[\arctan\left(\frac{1}{a+b}\right)+\arctan\left(\frac{1}{a+c}\right)=\arctan\left(\frac{1}{a}\right).\]
Do we know what \(\tan(A+B)\) is?
Could we use \(P=\arctan\left(\dfrac{1}{a+b}\right) \implies \tan P=\dfrac{1}{a+b}\)?
Can we simplify the LHS of the first equation with this?
The positive numbers \(p\), \(q\), \(r\), \(s\), \(t\), \(u\) and \(v\) satisfy \[st=(p+q)^2+1, \qquad uv=(p+r)^2+1, \qquad qr=p^2+1.\]
Prove that
\[\begin{align*} \arctan\left(\frac{1}{p+q+s}\right)&+\arctan\left(\frac{1}{p+q+t}\right)+\arctan\left(\frac{1}{p+r+u}\right)\\ &+\arctan\left(\frac{1}{p+r+v}\right)=\arctan\left(\frac{1}{p}\right). \end{align*}\]Can we use the first part of the question? Could we use it twice?
Hence show that \[\arctan\left(\frac{1}{13}\right)+\arctan\left(\frac{1}{21}\right)+\arctan\left(\frac{1}{82}\right)+\arctan\left(\frac{1}{187}\right)=\arctan\left(\frac{1}{7}\right).\]
Can we use simultaneous equations to find suitable values for the various letters?
