Inverse Circular Functions 2 Question 2
2. The value of cot $\sum _{n=1}^{19} \cot ^{-1} 1+\sum _{p=1}^{n} 2 p \quad$ is
(a) $\frac{23}{22}$
(b) $\frac{21}{19}$
(c) $\frac{19}{21}$
(d) $\frac{22}{23}$
(2019 Main, 10 Jan II)
Show Answer
Answer:
Correct Answer: 2. (b)
Solution:
- Consider, $\cot \sum _{n=1}^{19} \cot ^{-1} 1+\sum _{p=1}^{n} 2 p$
$$ \begin{aligned} & =\cot \sum _{n=1}^{19} \cot ^{-1}(1+n(n+1)) \quad \because \sum _{n=1}^{19} p=\frac{n(n+1)}{2} \\ & =\cot \sum _{n=1}^{19} \cot ^{-1}\left(1+n+n^{2}\right) \\ & =\cot \sum _{n=1}^{19} \tan ^{-1} \frac{1}{1+n(n+1)} \end{aligned} $$
$$ \left[\because \cot ^{-1} x=\tan ^{-1} \frac{1}{x}, \text { if } x>0\right] $$
$=\cot \sum _{n=1}^{19} \tan ^{-1} \frac{(n+1)-n}{1+n(n+1)} \quad[$ put $1=(n+1)-n]$
$=\cot \sum _{n=1}^{19}\left(\tan ^{-1}(n+1)-\tan ^{-1} n\right)$
$$ \because \tan ^{-1} \frac{x-y}{1+x y}=\tan ^{-1} x-\tan ^{-1} y $$
$=\cot \left[\left(\tan ^{-1} 2-\tan ^{-1} 1\right)+\left(\tan ^{-1} 3-\tan ^{-1} 2\right)+\right.$
$=\cot \left(\tan ^{-1} 20-\tan ^{-1} 1\right)$ .. $\left.+\left(\tan ^{-1} 20-\tan ^{-1} 19\right)\right]$
$=\cot \frac{\pi}{2}-\cot ^{-1} 20-\frac{\pi}{2}-\cot ^{-1} 1$
$$ \left[\because \tan ^{-1} x+\cot ^{-1} x=\pi / 2\right] $$
$=\cot \left(\cot ^{-1} 1-\cot ^{-1} 20\right)$
$=\frac{\cot \left(\cot ^{-1} 1\right) \cot \left(\cot ^{-1} 20\right)+1}{\cot \left(\cot ^{-1} 20\right)-\cot \left(\cot ^{-1} 1\right)}$
$=\frac{(1 \times 20)+1}{20-1}$
$$ \left[\because \cot (A-B)=\frac{\cot A \cot B+1}{\cot B-\cot A}\right. $$
$=\frac{21}{19}$ $\left[\because \cot \left(\cot ^{-1} x\right)=x\right]$
