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Question

the value of sinπ7+sin2π7+sin3π7 is equal to :-

A
cotπ14
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B
12cotπ14
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C
tanπ14
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D
12tanπ14
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Solution

The correct option is C 12cotπ14
sinπ7+sin2π7+sin3π7
2sin3π14cos(π4)+2sin3π14cos3π14
{sinC+sinD=2sinC+D2cosCD2
sin2x=2sinxcosx}
{cos(θ)=cosθ}
=2sin3π14(cosπ14+cos3π14)
=2sin3π14[2cosπ7cosπ14]
=4cos(π23π14)cos2π14cosπ14 {sinθ=cos(π2θ)}
=4cos4π14.cos2π14.cosπ14×sinπ14sinπ14
=2cos4π14.cos2π14.sin2π14sinπ14 {sin2θ=2sinθcosθ}
=cos4π14.sin4π14sinπ14×22
=sin8π142sinπ14=cos(π28π14)2sinπ14=cosπ142sinπ14=12cotπ14
Option (B) is correct.

1057742_1180981_ans_ea413003cb53451fbbd39b8d58e29814.jpg

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