Oxford University MAT 2020 Solved Papers

The geometric sum converges if , and converges to
$\frac{1}{\tan \mathit{x}}+\frac{1}{{\tan }^2\mathit{x}}+\frac{1}{{\tan }^3\mathit{x}}+\cdots =\frac{\frac{1}{\tan \mathit{x}}}{1-\frac{1}{\tan \mathit{x}}}=\frac{1}{\tan \mathit{x}-1}.$
Let $\mathit{u}=\tan \mathit{x}$, then this sum is equal to $\tan \mathit{x}$ if and only if ${\mathit{u}}^2-\mathit{u}=1$, so if and only if $\mathit{u}=\frac{1}{2}\left(1\pm \sqrt{5}\right)$. But the condition $\left|\tan \mathit{x}\right|>1$ means that we must take $\mathit{u}=\frac{1}{2}\left(1+\sqrt{5}\right)$. There is one value of $\mathit{x}$ in the range with this value of $\tan \mathit{x}$.