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यदि ${a_1},\;{a_2},............,{a_n}$ एक समांतर श्रेणी में हैं, जिसका सार्वान्तर $d$ है, तब श्रेणी $\sin d(\cos {\rm{ec}}\,{a_1}.{\rm{cosec}}\,{a_2} + {\rm{cosec}}\,{a_2}.{\rm{cosec}}\,{a_3} + ...........$ $ + {\rm{cosec}}\;{a_{n - 1}}{\rm{cosec}}\;{a_{n - 1}}{\rm{cosec}}\;{a_n})$
$\sec {a_1} - \sec {a_n}$
$\cot {a_1} - \cot {a_n}$
$\tan {a_1} - \tan {a_n}$
$c{\rm{osec}}\;{a_1} - {\rm{cosec}}\;{a_n}$
Solution
(b) दिये अनुसार, $d = {a_2} – {a_1} = {a_3} – {a_2} = …. = {a_n} – {a_{n – 1}}$
$\therefore $ $\sin d\,\{ {\rm{cosec}}\;{a_1}{\rm{cosec}}\;{a_2} + ….. + {\rm{cosec}}\;{a_{n – 1}}{\rm{cosec}}\;{a_n}\} $
$ = \frac{{\sin ({a_2} – {a_1})}}{{\sin {a_1}.\;\sin {a_2}}} + …… + \frac{{\sin ({a_n} – {a_{n – 1}})}}{{\sin {a_{n – 1}}\sin {a_n}}}$
$ = (\cot {a_1} – \cot {a_2}) + (\cot {a_2} – \cot {a_3}) + ….$
$ + (\cot {a_{n – 1}} – \cot {a_n})$
$ = \cot {a_1} – \cot {a_n}$.
