JEE PYQ: Inverse Trigonometric Functions Question 1
Question 1 - 2021 (16 Mar Shift 1)
Let $S_k = \sum_{r=1}^{k} \tan^{-1}\left(\frac{6^r}{2^{2r+1} + 3^{2r+1}}\right)$. Then $\lim_{k \to \infty} S_k$ is equal to:
(1) $\tan^{-1}\left(\frac{3}{2}\right)$
(2) $\frac{\pi}{2}$
(3) $\cot^{-1}\left(\frac{3}{2}\right)$
(4) $\tan^{-1}(3)$
Show Answer
Answer: (3)
Solution
Divide by $3^{2r}$. Rewrite as telescoping sum: $\sum_{r=1}^{k} \left[\tan^{-1}\left(\frac{2}{3}\right)^r - \tan^{-1}\left(\frac{2}{3}\right)^{r+1}\right]$. So $S_k = \tan^{-1}\left(\frac{2}{3}\right) - \tan^{-1}\left(\frac{2}{3}\right)^{k+1}$. As $k \to \infty$, $S_\infty = \tan^{-1}\left(\frac{2}{3}\right) - \tan^{-1}(0) = \tan^{-1}\left(\frac{2}{3}\right) = \cot^{-1}\left(\frac{3}{2}\right)$.
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