Question

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

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

Answer 1

Answer: (B)

We have, cot $\left[\sum _{n=1}^{23}co{t}^{-1}\right(1+\sum _{k=1}^{n}2k\left)\right]$
$\Rightarrow$ cot $[\sum _{n=1}^{23}co{t}^{-1}(1+2+4+6+8+......+2n)]$
$\Rightarrow$ cot $[\sum _{n=1}^{23}co{t}^{-1}\{1+n(n+1)\}]$
$\Rightarrow$ cot $\left[\sum _{n=1}^{23}ta{n}^{-1}\frac{1}{1+n(n+1)}\right]$
$\Rightarrow$ cot $\left[\sum _{n=1}^{23}ta{n}^{-1}\right\{\frac{(n+1)-n}{1+n(n+1)}\left\}\right]$
$\Rightarrow$ cot $[\sum _{n=1}^{23}(ta{n}^{-1}(n+1)-ta{n}^{-1}lnn)]$
$\Rightarrow$ cot $[(ta{n}^{-1}2-ta{n}^{-1}1)+(ta{n}^{-1}3-ta{n}^{-1}2)]$
$+(ta{n}^{-1}4-ta{n}^{-1}3)]+...+(ta{n}^{-1}24-ta{n}^{-1}23)]$
$\Rightarrow$ cot $(ta{n}^{-1}24-ta{n}^{-1}1)$
$\Rightarrow$ cot $\left(ta{n}^{-1}\frac{24-1}{1+24.(1)}\right)=cot\left(ta{n}^{-1}\frac{23}{25}\right)$
$=cot\left(co{t}^{-1}\frac{25}{23}\right)=\frac{25}{23}$