If operatorname{Tan}^{-1}left[frac{1}{1+1(2)} | vedclass

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(D) We know that $\operatorname{Tan}^{-1} x - \operatorname{Tan}^{-1} y = \operatorname{Tan}^{-1} \left( \frac{x-y}{1+xy} \right)$.Each term in the se

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$\frac{n}{n+2}$

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(D) We know that $\operatorname{Tan}^{-1} x - \operatorname{Tan}^{-1} y = \operatorname{Tan}^{-1} \left( \frac{x-y}{1+xy} ight)$.Each term in the series can be written as $\operatorname{Tan}^{-1} \left( \frac{(k+1)-k}{1+k(k+1)} ight) = \operatorname{Tan}^{-1}(k+1) - \operatorname{Tan}^{-1}(k)$.Summing from $k=1$ to $n$:$S = \sum_{k=1}^{n} (\operatorname{Tan}^{-1}(k+1) - \operatorname{Tan}^{-1}(k))$$S = (\operatorname{Tan}^{-1} 2 - \operatorname{Tan}^{-1} 1) + (\operatorname{Tan}^{-1} 3 - \operatorname{Tan}^{-1} 2) + \cdots + (\operatorname{Tan}^{-1}(n+1) - \operatorname{Tan}^{-1} n)$This is a telescoping series,so all intermediate terms cancel out:$S = \operatorname{Tan}^{-1}(n+1) - \operatorname{Tan}^{-1}(1)$Using the formula $\operatorname{Tan}^{-1} x - \operatorname{Tan}^{-1} y = \operatorname{Tan}^{-1} \left( \frac{x-y}{1+xy} ight)$:$S = \operatorname{Tan}^{-1} \left( \frac{(n+1)-1}{1+(n+1)(1)} ight) = \operatorname{Tan}^{-1} \left( \frac{n}{1+n+1} ight) = \operatorname{Tan}^{-1} \left( \frac{n}{n+2} ight)$.Comparing this with $\operatorname{Tan}^{-1} 	heta$,we get $	heta = \frac{n}{n+2}$.

If $\operatorname{Tan}^{-1}\left[\frac{1}{1+1(2)}\right]+\operatorname{Tan}^{-1}\left[\frac{1}{1+(2)(3)}\right]+\operatorname{Tan}^{-1}\left[\frac{1}{1+(3)(4)}\right]+\cdots+\operatorname{Tan}^{-1}\left[\frac{1}{1+n(n+1)}\right]=\operatorname{Tan}^{-1} \theta$,then $\theta=$

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