$\cot \left[\sum_{n=3}^{32} \cot ^{-1}\left(1+\sum_{k=1}^n 2 k\right)\right]=$

Question

Aicharya · Accepted Answer

$\begin{aligned} & \because \sum_{k=1}^n 2 k=n(n+1) \ & \therefore \cot \left(\sum_{n=3}^{32} \cot ^{-1}\left(1+\sum_{k=1}^n 2 k\right)\right) \ & =\cot \left(\sum_{n=3}^{32} \cot ^{-1}(1+n(n+1))\right) \ & =\cot \left(\sum_{n=3}^{32} \tan ^{-1}\left(\frac{(n+1)-n}{1+(n+1) n}\right)\right) \ & \because \sum_{n=3}^{32} \tan ^{-1}\left(\frac{(n+1)-n}{1+(n+1) n}\right)=\sum_{n=3}^{32}\left[\tan ^{-1}(n+1)-\tan ^{-1} n\right] \ & =\left(\tan ^{-1} 4-\tan ^{-1} 3+\left(\tan ^{-1} 5-\tan ^{-1} 4\right)+\ldots . . +\left(\tan ^{-1} 33-\tan ^{-1} 34\right)\right. \ & =\tan ^{-1} 33-\tan ^{-1} 3=\tan ^{-1}\left[\frac{33-3}{1+99}\right]=\tan ^{-1}\left(\frac{3}{10}\right) \ & \therefore \cot \left(\sum_{n=3}^{32} \tan ^{-1} \frac{(n+1)-n}{1+(n+1) n}\right)=\cot ^{-1}\left[\tan ^{-1}\left(\frac{3}{10}\right)\right] \ & =\cot ^{-1}\left[\cot ^{-1}\left(\frac{10}{3}\right)\right]=\frac{10}{3} \end{aligned}$ Hence, option (1) is correct.

\(\cot \left[\sum_{n=3}^{32} \cot ^{-1}\left(1+\sum_{k=1}^n 2 k\right)\right]=\)

Solution