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:
- 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} $
