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Question

In ΔABC, sinAcosBcosC+sinBcosCcosA+sinCcosAcosB is equal to


A
cosAcosBcosC
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B
sinAsinBsinC
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C
0
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D
None of these
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Solution

The correct option is B sinAsinBsinC
In ABC,A+B+C=π

A+B=πC ...... (i)

Consider,

sinAcosBcosC+sinBcosCcosA+sinCcosAcosB

=cosC(sinAcosB+cosAsinB)+sinCcosA.cosB

=cosC.sin(A+B)+sinC.cosA.cosB

=cosC.sin(πC)+sinC.cosA.cosB ...... From (i)

=sinC.cosC+sinC.cosAcosB

=sinC[cosC+cosA.cosB]

=sinC[cos(π(A+B))+cosA.cosB]

=sinC[cosA.cosBcos(A+B)]

=sinC[cosA.cosB(cosA.cosBsinA.sinB)]

=sinC[sinA.sinB]

=sinA.sinB.sinC.

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