JEE PYQ: Inverse Trigonometric Functions Question 3
Question 3 - 2021 (17 Mar Shift 1)
If $\cot^{-1}(\alpha) = \cot^{-1} 2 + \cot^{-1} 8 + \cot^{-1} 18 + \cot^{-1} 32 + \ldots$ upto 100 terms, then $\alpha$ is:
(1) 1.01
(2) 1
(3) 1.02
(4) 1.03
Show Answer
Answer: (1)
Solution
$\cot^{-1}(\alpha) = \sum_{r=1}^{100} \tan^{-1}\left(\frac{1}{2r^2}\right) = \sum_{r=1}^{100} \left[\tan^{-1}(2r+1) - \tan^{-1}(2r-1)\right] = \tan^{-1} 201 - \tan^{-1} 1 = \tan^{-1}\left(\frac{200}{202}\right) = \tan^{-1}\left(\frac{100}{101}\right)$. Therefore $\cot^{-1}(\alpha) = \cot^{-1}\left(\frac{101}{100}\right)$, so $\alpha = 1.01$.
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