Question
If $${a_1},{a_2},{a_3},.....,{a_n}$$ is an A.P. with common difference $$d;\left( {d > 0} \right)$$ then $$\tan \left[ {{{\tan }^{ - 1}}\left( {\frac{d}{{1 + {a_1}{a_2}}}} \right) + {{\tan }^{ - 1}}\left( {\frac{d}{{1 + {a_2}{a_3}}}} \right) + ..... + {{\tan }^{ - 1}}\left( {\frac{d}{{1 + {a_{n - 1}}{a_n}}}} \right)} \right]$$ is equal to
A.
$$\frac{{\left( {n - 1} \right)d}}{{{a_1} + {a_n}}}$$
B.
$$\frac{{\left( {n - 1} \right)d}}{{1 + {a_1}{a_n}}}$$
C.
$$\frac{{nd}}{{1 + {a_1}{a_n}}}$$
D.
$$\frac{{{a_n} - {a_1}}}{{{a_n} + {a_1}}}$$
Answer :
$$\frac{{\left( {n - 1} \right)d}}{{1 + {a_1}{a_n}}}$$
Solution :
We have,
$$\eqalign{
& {\tan ^{ - 1}}\left( {\frac{d}{{1 + {a_1}{a_2}}}} \right) + {\tan ^{ - 1}}\left( {\frac{d}{{1 + {a_2}{a_3}}}} \right) + ..... + {\tan ^{ - 1}}\left( {\frac{d}{{1 + {a_{n - 1}}{a_n}}}} \right) \cr
& = {\tan ^{ - 1}}\left( {\frac{{{a_2} - {a_1}}}{{1 + {a_1}{a_2}}}} \right) + {\tan ^{ - 1}}\left( {\frac{{{a_3} - {a_2}}}{{1 + {a_2}{a_3}}}} \right) + ..... + {\tan ^{ - 1}}\left( {\frac{{{a_n} - {a_{n - 1}}}}{{1 + {a_{n - 1}}{a_n}}}} \right) \cr
& = \left( {{{\tan }^{ - 1}}{a_2} - {{\tan }^{ - 1}}{a_1}} \right) + \left( {{{\tan }^{ - 1}}{a_3} - {{\tan }^{ - 1}}{a_2}} \right) + ..... + \left( {{{\tan }^{ - 1}}{a_n} - {{\tan }^{ - 1}}{a_{n - 1}}} \right) \cr
& = {\tan ^{ - 1}}{a_n} - {\tan ^{ - 1}}{a_1} = {\tan ^{ - 1}}\left( {\frac{{{a_n} - {a_1}}}{{1 + {a_n}{a_1}}}} \right) \cr
& = {\tan ^{ - 1}}\left( {\frac{{\left( {n - 1} \right)d}}{{1 + {a_1}{a_n}}}} \right) \cr
& \therefore \tan \left[ {{{\tan }^{ - 1}}\left( {\frac{d}{{1 + {a_1}{a_2}}}} \right) + {{\tan }^{ - 1}}\left( {\frac{d}{{1 + {a_2}{a_3}}}} \right) + ..... + {{\tan }^{ - 1}}\left( {\frac{d}{{1 + {a_{n - 1}}{a_n}}}} \right)} \right] = \frac{{\left( {n - 1} \right)d}}{{1 + {a_1}{a_n}}} \cr} $$