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Question

$I f, csc A + csc B + csc C = 0$ $If,cscA+cscB+cscC=0$ ,then show that ${(\sum sin A)}^{2} = \sum {sin}^{2} A$ $(sumsinA)^2=sumsin^2A$

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Answer 1 · 2018-03-23T07:46:38

Please see below.

${(\sum sin A)}^{2} = {(sin A + sin B + sin C)}^{2}$ $(sumsinA)^2=(sinA+sinB+sinC)^2$

= ${sin}^{2} A + {sin}^{2} B + {sin}^{2} C + 2 sin A sin B + 2 sin B sin C + 2 sin C sin A$ $sin^2A+sin^2B+sin^2C+2sinAsinB+2sinBsinC+2sinCsinA$

= $\sum {sin}^{2} A + 2 sin A sin B + 2 sin B sin C + 2 sin C sin A$ $sumsin^2A+2sinAsinB+2sinBsinC+2sinCsinA$

= $\sum {sin}^{2} A + 2 sin A sin B sin C (\frac{1}{sin C} + \frac{1}{sin B} + \frac{1}{sin A})$ $sumsin^2A+2sinAsinBsinC(1/sinC+1/sinB+1/sinA)$

= $\sum {sin}^{2} A + 2 sin A sin B sin C (csc C + csc B + csc A)$ $sumsin^2A+2sinAsinBsinC(cscC+cscB+cscA)$

= $\sum {sin}^{2} A + 2 sin A sin B sin C \times 0$ $sumsin^2A+2sinAsinBsinCxx0$

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