If a_1, a_2, a_3, dots, a_n are in arithmetic | vedclass

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(C) We know that $	an^{-1} x - 	an^{-1} y = 	an^{-1} (\frac{x - y}{1 + xy})$.Since $a_1, a_2, \dots, a_n$ are in arithmetic progression, $a_{k+1} -

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$\frac{a_n - a_1}{1 + a_1a_n}$

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(C) We know that $	an^{-1} x - 	an^{-1} y = 	an^{-1} (\frac{x - y}{1 + xy})$.Since $a_1, a_2, \dots, a_n$ are in arithmetic progression, $a_{k+1} - a_k = d$.Thus, each term in the sum can be written as $	an^{-1} (\frac{a_{k+1} - a_k}{1 + a_k a_{k+1}}) = 	an^{-1} a_{k+1} - 	an^{-1} a_k$.The given expression becomes $	an [(	an^{-1} a_2 - 	an^{-1} a_1) + (	an^{-1} a_3 - 	an^{-1} a_2) + \dots + (	an^{-1} a_n - 	an^{-1} a_{n-1})]$.This is a telescoping sum, which simplifies to $	an [	an^{-1} a_n - 	an^{-1} a_1]$.Using the formula $	an (	an^{-1} x - 	an^{-1} y) = \frac{x - y}{1 + xy}$, we get $\frac{a_n - a_1}{1 + a_n a_1}$.

If $a_1, a_2, a_3, \dots, a_n$ are in arithmetic progression with common difference $d$, then $\tan [\tan^{-1} (\frac{d}{1 + a_1a_2}) + \tan^{-1} (\frac{d}{1 + a_2a_3}) + \dots + \tan^{-1} (\frac{d}{1 + a_{n-1}a_n})] = $

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