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Question

cosAcsinB+cosBasinC+cosCbsinA=

A
Rabc
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B
2R3abc
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C
1R
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D
R3abc
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Solution

The correct option is C 1R
In ABC
cosAcsinB+cosBasinC+cosCbsinA=?
Using the lae of sines,
asinA=bsinB=csinC=Ra=2RsinAb=2RsinBc=2RsinCcosA2RsinCsinB+cosB2RsinAsinC+cosC2RsinBsinA=2cosA.sinA4RsinAsinBsinC+2cosBsinC4RsinAsinBsinC+2cosCsinC4RsinAsinBsinC=sin2A+sin2B+sin2C4RsinAsinBsinC
In a ABC, we have an identity,
sin2A+sin2B+sin2C=4sinAsinBsinC
Using it we can reduce,
=4sinAsinBsinC4RsinAsinBsinC=1RcosAcsinB+cosBasinC+cosCbsinA=1R

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