Question

The value of $cot\left(\sum _{n=1}^{23}co{t}^{-1}\left(1+\sum _{k=1}^{n}2k\right)\right)$ is

• A. $\frac{23}{25}$
• B. $\frac{25}{23}$
• C. $\frac{23}{24}$
• D. $\frac{24}{23}$

Answer 1

The correct option is B

$\frac{25}{23}$

Explanation for the correct option:

Find the required value.

We have,

$$\begin{gathered} cot\left(\sum _{n=1}^{23}co{t}^{-1}\left(1+\sum _{k=1}^{n}2k\right)\right)=cot\left(\sum _{n=1}^{23}co{t}^{-1}\left(1+2\times {\displaystyle \frac{n(n+1)}{2}}\right)\right) \\ =cot\left(\sum _{n=1}^{23}co{t}^{-1}\left(1+n(n+1)\right)\right) \\ =cot\left(\sum _{n=1}^{23}{\tan }^{-1}\left({\displaystyle \frac{1}{1+n(n+1)}}\right)\right) \\ =cot\left(\sum _{n=1}^{23}{\tan }^{-1}\left({\displaystyle \frac{n+1-n}{1+n(n+1)}}\right)\right) \\ =cot\left(\sum _{n=1}^{23}{\tan }^{-1}\left(n+1\right)-{\tan }^{-1}n\right)\left[\because {\tan }^{-1}A-{\tan }^{-1}B={\tan }^{-1}\left({\displaystyle \frac{A-B}{1+AB}}\right)\right] \\ =cot\left({\tan }^{-1}24-{\tan }^{-1}1\right) \\ =cot\left({\tan }^{-1}\left({\displaystyle \frac{24-1}{1+24\left(1\right)}}\right)\right) \\ =cot\left({\tan }^{-1}\left({\displaystyle \frac{23}{25}}\right)\right) \\ =cot\left(co{t}^{-1}\left({\displaystyle \frac{25}{23}}\right)\right) \\ =\frac{25}{23} \end{gathered}$$

Hence, option (B) is the correct answer.