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Question

In a triangle ABC, if b2+c2=3a2, then cotB+cotC=

A
2cRac
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B
2aRc2
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C
2aRbc
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D
None of these
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Solution

The correct option is C 2aRbc
b2+c2a2=2a2

2bccosA=2a2

cosA=2a22bc=a2bc=sin2AsinBsinC

cotB+cotC=cosBsinB+cosCsinC

=sinCcosB+cosCsinBsinBsinC

=sin(B+C)sinBsinC

=sin(πA)sinBsinC since A+B+C=π

=sinAsinBsinC

Using sine rule

asinA=bsinB=csinC=2R

sinA=a2R,sinB=b2R,sinC=c2R

sinAsinBsinC

=a2Rb2Rc2R

=2Rabc

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