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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) $

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  • \( d=a_2-a_1=a_3-a_2=a_4-a_3=......a_n-a_{n-1}\)
  • \( tan^{-1}\frac{x-y}{1+xy}=tan^{-1}x-tan^{-1}y\:\:\:xy<1\)
\(Since a_1,a_2,a_3...........a_n\) 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
\( tan \bigg[ \bigg( tan^{-1}a_2 - tan^{-1}a_1 \bigg) + (tan^{-1}a_3+tan^{-1}a_2 \bigg) +...... \bigg( tan^{-1}a_n-tan^{-1}a_{n-1} \bigg) \bigg] \)
\( = tan \bigg[ tan^{-1}a_n-tan^{-1}a_1 \bigg] \)
\( = tan \bigg( tan^{-1} \frac{a_n-a_1}{1+a_1a_n} \bigg) = \frac{a_n-a_1}{1+a_1a_n} \)


answered Feb 18, 2013 by thanvigandhi_1
edited Mar 15, 2013 by rvidyagovindarajan_1

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