Question

Q.63:


Q63 The value of $cot\left(\sum _{n=1}^{23}co{t}^{-1}\left(1+\sum _{k=1}^{n}2k\right)\right)$

(A) 23/25 (B) 25/23 (C) 23/24 (D) 24/23

Answer 1

Dear student
$$\begin{gathered} \cot \left(\sum _{\mathrm{n}=1}^{23}{\cot }^{-1}\left(1+\sum _{\mathrm{k}=1}^{\mathrm{n}}2\mathrm{k}\right)\right) \\ =\cot \left(\sum _{\mathrm{n}=1}^{23}{\cot }^{-1}\left(1+\mathrm{n}\left(\mathrm{n}+1\right)\right)\right)\left[\because \sum _{\mathrm{k}=1}^{\mathrm{n}}2\mathrm{k}=\mathrm{n}(\mathrm{n}+1)\right] \\ =\cot \left(\sum _{\mathrm{n}=1}^{23}{\cot }^{-1}\left(1+{\mathrm{n}}^2+\mathrm{n}\right)\right) \\ =\cot \left(\sum _{\mathrm{n}=1}^{23}{\tan }^{-1}\frac{1}{{\mathrm{n}}^2+\mathrm{n}+1}\right)\left[\because {\cot }^{-1}\mathrm{x}={\tan }^{-1}\frac{1}{\mathrm{x}}\right] \\ =\cot \left(\sum _{\mathrm{n}=1}^{23}{\tan }^{-1}\frac{\mathrm{n}+1-\mathrm{n}}{1+\mathrm{n}(\mathrm{n}+1)}\right)\left[\because {\tan }^{-1}\mathrm{x}-{\tan }^{-1}\mathrm{y}={\tan }^{-1}\left(\frac{\mathrm{x}-\mathrm{y}}{1+\mathrm{xy}}\right)\right] \\ =\cot \left(\sum _{\mathrm{n}=1}^{23}{\tan }^{-1}(\mathrm{n}+1)-{\tan }^{-1}\mathrm{n}\right) \\ =\cot \left({\tan }^{-1}24-{\tan }^{-1}1\right) \\ =\cot \left({\tan }^{-1}\left(\frac{24-1}{1+24\left(1\right)}\right)\right)\left[\because {\tan }^{-1}\mathrm{x}-{\tan }^{-1}\mathrm{y}={\tan }^{-1}\left(\frac{\mathrm{x}-\mathrm{y}}{1+\mathrm{xy}}\right)\right] \\ =\cot \left({\tan }^{-1}\left(\frac{23}{25}\right)\right) \\ =\cot \left({\cot }^{-1}\left(\frac{25}{23}\right)\right)\left[\because {\tan }^{-1}\mathrm{x}={\cot }^{-1}\left(\frac{1}{\mathrm{x}}\right)\right] \\ =\frac{25}{23} \end{gathered}$$
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