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# If $a_1,a_2,a_3,......a_n$ is an arithmetic progression with common difference d,then evaluate the following expression $\tan \bigg( \tan^{-1}\bigg(\frac{d}{1+a_1a_2}\bigg)+\tan^{-1}\bigg(\frac{d}{1+a_2a_3}\bigg)+\tan^{-1}\bigg(\frac{d}{1+a_3a_4}\bigg)+........+\tan^{-1}\bigg(\frac{d}{1+a_{n-1}a_n}\bigg) \bigg)$

Toolbox:
• $d=a_2-a_1=a_3-a_2=a_4-a_3=......a_n-a_{n-1}$
• $tan^{-1}\frac{x-y}{1+xy}=tan^{-1}x-tan^{-1}y\:\:\:xy<1$
$Since a_1,a_2,a_3...........a_n$ are in A.P. we know that
$d=a_2-a_1=a_3-a_2=a_4-a_3=......a_n-a_{n-1}$
By taking $d=a_2-a_2$, we get $tan^{-1} \bigg( \frac{a_2-a_1}{1+a_1a_2} \bigg) = tan^{-1}a_2-tan^{-1}a_1$
Substituting the values we get
$tan^{-1} \bigg( \frac{d}{1+a_1a_2} \bigg) = tan^{-1} \bigg( \frac{a_2-a_1}{1+a_1a_2} \bigg) = tan^{-1}a_2-tan^{-1}a_1$
Similarly by taking $d=a_3-a_2=a_4-a_3=...........we\:get$

$tan^{-1} \frac{d}{1+a_2a_3}=tan^{-1} \bigg( \frac{a_3-a_2}{1+a_2a_3} \bigg).....$
$\Rightarrow$ The given expression becomes

$tan \bigg[ \bigg( tan^{-1}a_2 - tan^{-1}a_1 \bigg) + (tan^{-1}a_3+tan^{-1}a_2 \bigg) +...... \bigg( tan^{-1}a_n-tan^{-1}a_{n-1} \bigg) \bigg]$
$= tan \bigg[ tan^{-1}a_n-tan^{-1}a_1 \bigg]$
$= tan \bigg( tan^{-1} \frac{a_n-a_1}{1+a_1a_n} \bigg) = \frac{a_n-a_1}{1+a_1a_n}$

edited Mar 15, 2013