Let \(A(x)=\arctan(x)\). Further, for any real \(x\), let \[ u_n(x)=\frac{\alpha(x)^n+\beta(x)^n}{2}\quad\text{and}\quad v_n(x)=\frac{\alpha(x)^n-\beta(x)^n}{\alpha-\beta}, \] where \((\alpha(x),\beta(x))=(1+ix, 1-ix)\), \(i={\sqrt{-1}}\). The authors show that for any positive integer \(n\) and any real number \(x\neq \pm 1\), we have \(A^{n}(1/x)=v_n(x)/u_n(x)\) (Theorem 1.3) and \(A^n(x)=(-1)^{n+1} (v_n(x)/u_n(x))^{(-1)^n}\) (Theorem 1.4). As a consequence, they exhibit identities such as \[ n\arctan \left(\frac{1}{x}\right)+\arctan\left(\frac{u_n(x)-v_n(x)}{u_n(x)+v_n(x)}\right)=\frac{\pi}{4}+k(n,x)\pi, \] where \(k(n,x)\) is some suitable integer. Similarly, they prove that \[ 2\arctan (-x\pm {\sqrt{1+x^2}})+\arctan(x)=\pm \frac{\pi}{2} \] holds for every real number \(x\). Taking \(x=L_{2k+1}/2\), where \(\{L_m\}_{m\geq 0}\) is the Lucas companion of the Fibonacci sequence, they get formulas such as \[ \frac{\pi}{2}=\arctan \left(\frac{L_{2k+1}}{2}\right)+2\arctan\left(\frac{1}{\phi^{2k+1}}\right), \] where \(\phi=(1+{\sqrt{5}})/2\) is the Golden mean. The proofs are by induction.