JEE PYQ: Inverse Trigonometric Functions Question 13
Question 13 - 2020 (05 Sep Shift 1)
If S is the sum of the first 10 terms of the series $\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{13}\right) + \tan^{-1}\left(\frac{1}{21}\right) + \ldots$, then $\tan(S)$ is equal to:
(1) $\frac{5}{6}$
(2) $\frac{5}{11}$
(3) $-\frac{6}{5}$
(4) $\frac{10}{11}$
Show Answer
Answer: (1)
Solution
General term: $\tan^{-1}\left(\frac{1}{r^2+r+1}\right) = \tan^{-1}(r+1) - \tan^{-1}(r)$ (telescoping). $S = \tan^{-1}(11) - \tan^{-1}(1) = \tan^{-1}\left(\frac{11-1}{1+11}\right) = \tan^{-1}\left(\frac{5}{6}\right)$. So $\tan(S) = \frac{5}{6}$.
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