Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

$cot \left(\underset{\text{n}=1}{\overset{23}{\Sigma }}{\text{cot}}^{-1}\left(1+\underset{\text{k}=1}{\overset{\text{n}}{\Sigma }}2\text{k}\right)\right)=\text{cot}\underset{\text{n}=1}{\overset{23}{\Sigma }}{\text{cot}}^{-1}\left(1+2+4+6+\dots +2\text{n}\right)$
$\text{=cot}\Sigma {\text{cot}}^{-1}\left(1+\text{n}\left(\text{n}+1\right)\right)$
$\text{=cot}\Sigma {\text{tan}}^{-1}\frac{\left(\text{n}+1\right)-\text{n}}{1+\text{n}\left(\text{n}+1\right)}$ $\left(co{t}^{-1}x={\tan }^{-1}\frac{1}{x}\right)$
$\text{=cot}\underset{\text{n}=1}{\overset{23}{\Sigma }}\left({\text{tan}}^{-1}\left(\text{n}+1\right)-{\text{tan}}^{-1}\text{n}\right)\left[{\text{tan}}^{\text{-1}}\left(\frac{\text{x-y}}{1+xy}\right){\text{=tan}}^{\text{-1}}{\text{x-tan}}^{\text{-1}}\text{y}\right]$
 

Now ${\text{tan}}^{-1}\left(\text{n}+1\right)-{\text{tan}}^{-1}{\text{n=tan}}^{-1}\left(2\right)-{\text{tan}}^{-1}{\text{1+tan}}^{-1}\left(3\right)-{\text{tan}}^{-1}{\text{2+....+tan}}^{-1}\left(24\right)-{\text{tan}}^{-1}\text{23}$

$\text{=cot}\left({\text{tan}}^{-1}24-{\text{tan}}^{-1}1\right)$
$\text{=cot}\left({\text{tan}}^{-1}\frac{24-1}{1+24}\right)$
$\text{=cot}\left({\text{cot}}^{-1}\frac{25}{23}\right)=\frac{25}{23}$