Fun with ArcTan

The key to this problem is the trig identity $\tan \left( A + B \right) = \frac{ \tan\left( A \right) + \tan \left( B \right) }{1- \tan\left( A \right) \tan\left( B \right) }.$ Since arctangent is the inverse of the tangent function, $\textcolor{#D61F06}{\tan} \left( \text{arctan} \left( \frac{1}{x} \right) + \text{arctan} \left( \frac{1}{x+2} \right) \right) = \textcolor{#D61F06}{\tan} \left( \text{arctan} \left( \frac{4}{x+3} \right) \right) = \frac{4}{x+3}.$ Using the trig identity, $\tan \left( \text{arctan} \left( \frac{1}{x} \right) + \text{arctan} \left( \frac{1}{x+2} \right) \right) = \frac{ \frac{1}{x} + \frac{1}{\left( x+2 \right)} }{1-\frac{1}{x}\frac{1}{\left( x+2 \right)}} = \frac{2 \left( x+1 \right)}{x^2+2x-1} = \frac{4}{x+3}$ after simplifying the complex fraction. After some algebra, we find $x^2-5=0,$ which has solutions $\pm \sqrt{5}$ . Only $\sqrt{5}$ is a valid solution, however.