JEE PYQ: Inverse Trigonometric Functions Question 19
Question 19 - 2019 (10 Jan Shift 2)
The value of $\cot\left(\sum_{n=1}^{19} \cot^{-1}\left(1 + \sum_{p=1}^{n} 2p\right)\right)$ is:
(1) $\frac{21}{19}$
(2) $\frac{19}{21}$
(3) $\frac{22}{23}$
(4) $\frac{23}{22}$
Show Answer
Answer: (1)
Solution
$\sum_{p=1}^{n} 2p = n(n+1)$, so $\cot^{-1}(1 + n(n+1))$. Now $\cot^{-1}(1+n(n+1)) = \tan^{-1}\left(\frac{(n+1)-n}{1+(n+1)n}\right) = \tan^{-1}(n+1) - \tan^{-1}(n)$. So $\sum_{n=1}^{19} = \tan^{-1}(20) - \tan^{-1}(1) = \tan^{-1}\left(\frac{19}{21}\right)$. Therefore $\cot\left(\tan^{-1}\frac{19}{21}\right) = \frac{21}{19}$.
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