If tan^{-1} frac{1}{1+1(2)} + tan^{-1} frac{1 | vedclass

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(C) The general term of the series is $T_k = 	an^{-1} \frac{1}{1+k(k+1)}$.Using the identity $	an^{-1} x - 	an^{-1} y = 	an^{-1} \frac{x-y}{1+xy}$

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

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(C) The general term of the series is $T_k = 	an^{-1} \frac{1}{1+k(k+1)}$.Using the identity $	an^{-1} x - 	an^{-1} y = 	an^{-1} \frac{x-y}{1+xy}$,we can write:$T_k = 	an^{-1} \frac{(k+1)-k}{1+k(k+1)} = 	an^{-1}(k+1) - 	an^{-1}(k)$.Summing from $k=1$ to $n$:$S_n = \sum_{k=1}^{n} (	an^{-1}(k+1) - 	an^{-1}(k))$$= (	an^{-1} 2 - 	an^{-1} 1) + (	an^{-1} 3 - 	an^{-1} 2) + \dots + (	an^{-1}(n+1) - 	an^{-1} n)$$= 	an^{-1}(n+1) - 	an^{-1} 1$.Using the formula $	an^{-1} x - 	an^{-1} y = 	an^{-1} \frac{x-y}{1+xy}$:$S_n = 	an^{-1} \frac{(n+1)-1}{1+(n+1)(1)} = 	an^{-1} \frac{n}{n+2}$.Given $S_n = 	an^{-1} 	heta$,we have $	heta = \frac{n}{n+2}$.

If $\tan^{-1} \frac{1}{1+1(2)} + \tan^{-1} \frac{1}{1+2(3)} + \tan^{-1} \frac{1}{1+3(4)} + \dots + \tan^{-1} \frac{1}{1+n(n+1)} = \tan^{-1} \theta$,then $\theta$ =

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