If $cosec A + cosec B + cosec C = 0$ , then show that ${(sin A + sin B + sin C)}^{2} = \sum {sin}^{2} A$

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Solution

$L H S = {(sin A + sin B + sin C)}^{2}$

$= {sin}^{2} A + {sin}^{2} B + {sin}^{2} C + 2 sin A sin B + 2 sin B sin C + 2 sin C sin A$

$= \sum {sin}^{2} A + 2 [\frac{1}{c o s e c A . c o s e c B} + \frac{1}{cos e c B cos e c C} + \frac{1}{c o s e c C . c o s e c A}]$

$= \sum {sin}^{2} A + 2 [\frac{c o s e c C + c o s e c B + c o s e c A}{c o s e c A . c o s e c B . c o s e c C}]$

$= \sum {sin}^{2} A + 2 [\frac{0}{cos e c A + cos e c B . cos e c C}]$ $(∵ cos e c A + cos e c B + cos e c C = 0)$

$= \sum {sin}^{2} A$

$= R H S$

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