Inverse Circular Functions 3 Question 7

7. The value of $\cot \sum _{n=1}^{23} \cot ^{-1} 1+\sum _{k=1}^{n} 2 k \quad$ is

(2013 Main)

(a) $\frac{23}{25}$

(b) $\frac{25}{23}$

(c) $\frac{23}{24}$

(d) $\frac{24}{23}$

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Answer:

Correct Answer: 7. (b)

Solution:

  1. We have, $\cot \sum _{n=1}^{23} \cot ^{-1} 1+\sum _{k=1}^{n} 2 k$

$\Rightarrow \cot \sum _{n=1}^{23} \cot ^{-1}(1+2+4+6+8+\ldots+2 n)$

$\Rightarrow \cot \sum _{n=1}^{23} \cot ^{-1}{1+n(n+1)}$

$\Rightarrow \cot \sum _{n=1}^{23} \tan ^{-1} \frac{1}{1+n(n+1)}$

$\Rightarrow \cot \sum _{n=1}^{23} \tan ^{-1} \frac{(n+1)-n}{1+n(n+1)}$

$\Rightarrow \cot \sum _{n=1}^{23}\left(\tan ^{-1}(n+1)-\tan ^{-1} \ln n\right)$

$\Rightarrow \cot \left[\left(\tan ^{-1} 2-\tan ^{-1} 1\right)+\left(\tan ^{-1} 3-\tan ^{-1} 2\right)\right.$

$ \left.\left.+\left(\tan ^{-1} 4-\tan ^{-1} 3\right)\right]+\ldots+\left(\tan ^{-1} 24-\tan ^{-1} 23\right)\right] $

$\Rightarrow \cot \left(\tan ^{-1} 24-\tan ^{-1} 1\right)$

$\Rightarrow \cot \tan ^{-1} \frac{24-1}{1+24 \cdot(1)}=\cot \tan ^{-1} \frac{23}{25}$

$ =\cot \cot ^{-1} \frac{25}{23}=\frac{25}{23} $



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