Question

Let $\theta ={\tan }^{-1}\left(\frac{1}{10}\right)+{\tan }^{-1}\left(\frac{1}{9}\right)+\dots \dots .+{\tan }^{-1}(1)+{\tan }^{-1}(2)+\dots \dots +{\tan }^{-1}(10)+{\tan }^{-1}(11)$. If $\tan \theta =\frac{p}{q}$ (where $p$ and $q$ are coprime), then find the value of $(p+q)$.

Answer 1

Answer: 11

Given $\theta ={\tan }^{-1}\left(\frac{1}{10}\right)+{\tan }^{-1}\left(\frac{1}{9}\right)+\dots \dots .+{\tan }^{-1}(1)+{\tan }^{-1}(2)+\dots \dots +{\tan }^{-1}(10)+{\tan }^{-1}(11)$
$\theta =\left({\tan }^{-1}\left(\frac{1}{10}\right)+{\tan }^{-1}(10)\right)+\left({\tan }^{-1}\left(\frac{1}{9}\right)+{\tan }^{-1}(9)\right)+\dots \dots .\left({\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}(2)\right)+{\tan }^{-1}(1)+{\tan }^{-1}(11)$
$\theta =9\left(\frac{\pi }{2}\right)+\frac{\pi }{4}+{\tan }^{-1}(11)=\frac{19\pi }{4}+{\tan }^{-1}(11)$
$\text{ Now }\tan \theta =\tan \left(\frac{19\pi }{4}+{\tan }^{-1}(11)\right)=\frac{11-1}{1+11}=\frac{10}{12}=\frac{5}{6}\equiv \frac{p}{q}\Rightarrow p+q=11.$