JEE PYQ: Inverse Trigonometric Functions Question 6
Question 6 - 2021 (24 Feb Shift 1)
$\lim_{n \to \infty} \tan\left{\sum_{r=1}^{n} \tan^{-1}\left(\frac{1}{1+r+r^2}\right)\right}$ is equal to:
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Answer: (1)
Solution
$\tan^{-1}\left(\frac{1}{1+r+r^2}\right) = \tan^{-1}(r+1) - \tan^{-1}(r)$. So $\sum_{r=1}^{n} = \tan^{-1}(n+1) - \tan^{-1}(1)$. As $n \to \infty$: $\frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$. Therefore $\tan\left(\frac{\pi}{4}\right) = 1$.
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