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  • if a1,a2,a3,.........................an is an arithmetic progression with common difference d, then evaluate the following expression tan into bracket tan inverse into bracket d divided 1 plus a1a2 plus tan inverse into bracket d divided 1 plus a2a3

If a1,a2,a3,.................an  is an arithmetic progression with common difference d, then evaluate the following expression.
\tan\left [ \tan^{-1}\left ( \frac{d}{1+a_{1}a_{2}} \right )+\tan^{-1}\left ( \frac{d}{1+a_{2}a_{3}} \right )+\tan^{-1}\left ( \frac{d}{1+a_{3}a_{4}} \right )+..........+\tan^{-1}\left ( \frac{d}{1+a_{n-1}a_{n}} \right ) \right ]
 

Answers (1)

We have a_{1}=a, a_{2}=a + d, a_{3}=a+2d.......

And, d=a_{2}-a_{1}=a_{3}-a_{2}=a_{4}-a_{3}=......=a_{n}-a_{n-1}

Given that,

\tan\left [ \tan^{-1}\left (\frac{d}{1+a_{1}a_{2}}\right )+ \tan^{-1}\left (\frac{d}{1+a_{2}a_{3}}\right )+ \tan^{-1}\left (\frac{d}{1+a_{3}a_{4}}\right )+............+ \tan^{-1}\left (\frac{d}{1+a_{n-1}a_{n}}\right ) \right ]

=\tan^{-1}\left [ \tan^{-1}\left ( \frac{a_{2}-a_{1}}{1+a_{1}a_{2}} \right )+\tan^{-1}\left ( \frac{a_{3}-a_{2}}{1+a_{2}a_{3}} \right )+.........+\tan^{-1}\left ( \frac{a_{n}-a_{n-1}}{1+a_{n-1}a_{n}} \right ) \right ]

=\tan\left [ \left ( \tan^{-1}a_{2}-\tan^{-1}a_{1} \right ) +\left ( \tan^{-1}a_{3}-\tan^{-1}a_{2} \right ) +................+\left ( \tan^{-1}a_{n}-\tan^{-1}a_{n-1} \right ) \right ]

=\tan \left [ \tan^{-1}a_{n}-\tan^{-1}a_{1} \right ]

\left [ Scince , \tan^{-1}x-\tan^{-1}y=\tan^{-1}\left ( \frac{x-y}{1+xy} \right ) \right ]

=\tan\left [ \tan^{-1} \left ( \frac{a_{n}-a_{1}}{1+a_{1}a_{n}} \right )\right ]

\left [ scince, \tan\left ( \tan^{-1}x \right )=x \right ]

=\frac{a_{n}-a_{1}}{1+a_{1}a_{n}}

 

 

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