We have, ${a_1} = a,{a_2} = a + d,{a_3} = a + 2d$
and $d = {a_2} – {a_1} = {a_3} – {a_2} = {a_4} – {a_3} = \ldots = {a_n} – {a_{n – 1}}$

If ${a_1},{a_2},{a_3}, \ldots ,{a_n}$ is an arithmetic progression with common difference
$d$, then evaluate the following Exemplarpression.
$\tan \left[ {{{\tan }^{ – 1}}\left( {\frac{d}{{1 + {a_1}{a_2}}}} \right) + {{\tan }^{ – 1}}\left( {\frac{d}{{1 + {a_2}{a_3}}}} \right) + {{\tan }^{ – 1}}\left( {\frac{d}{{1 + {a_3}{a_4}}}} \right)} \right.$ $\left. { + \ldots + {{\tan }^{ – 1}}\left( {\frac{d}{{1 + {a_{n – 1}}{a_n}}}} \right)} \right]$

We have, ${a_1} = a,{a_2} = a + d,{a_3} = a + 2d$
and $d = {a_2} – {a_1} = {a_3} – {a_2} = {a_4} – {a_3} = \ldots = {a_n} – {a_{n – 1}}$
[NCERT,Exemplar.2.3,Q.19,Page.37]

Given that, $\tan \left[ {{{\tan }^{ – 1}}\left( {\frac{d}{{1 + {a_1}{a_2}}}} \right) + {{\tan }^{ – 1}}\left( {\frac{d}{{1 + {a_2}{a_3}}}} \right)} \right.$.

$\left. { + {{\tan }^{ – 1}}\left( {\frac{d}{{1 + {a_3}{a_4}}}} \right) + \ldots + {{\tan }^{ – 1}}\left( {\frac{d}{{1 + {a_{n – 1}} \cdot {a_n}}}} \right)} \right]$

$ = \tan \left[ {{{\tan }^{ – 1}}\frac{{{a_2} – {a_1}}}{{1 + {a_2} \cdot {a_1}}} + {{\tan }^{ – 1}}\frac{{{a_3} – {a_2}}}{{1 + {a_3} \cdot {a_2}}} + \ldots + {{\tan }^{ – 1}}\frac{{{a_n} – {a_{n – 1}}}}{{1 + {a_n} \cdot {a_{n – 1}}}}} \right]$

$ = \tan \left[ {\left( {{{\tan }^{ – 1}}{a_2} – {{\tan }^{ – 1}}{a_1}} \right) + \left( {{{\tan }^{ – 1}}{a_3} – {{\tan }^{ – 1}}{a_2}} \right) + \ldots + \left( {{{\tan }^{ – 1}}{a_n} – {{\tan }^{ – 1}}{a_{n – 1}}} \right)} \right]$

$ = \tan \left[ {{{\tan }^{ – 1}}{a_n} – {{\tan }^{ – 1}}{a_1}} \right]$

$ = \tan \left[ {{{\tan }^{ – 1}}\frac{{{a_n} – {a_1}}}{{1 + {a_n} \cdot {a_1}}}} \right]$

$ = \frac{{{a_n} – {a_1}}}{{1 + {a_n} \cdot {a_1}}}$

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