Question

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

Answer 1

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