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Question

In a triangle ABC, if cosAa=tanCc, then sin(B+C) is equal to


A

cosBsinA

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B

cosAcosC

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C

cosAsinB

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D

sinBsinC

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Solution

The correct option is B

cosAcosC


Determine the value of sin(B+C)

Given, cosAa=tanCc

cosAa=tanCccosAa=sinCcosC×1ccosAa=sinCc×1cosCcosAa=sinAa×1cosC[fromSinerule]sinA=cosAcosCsin(π-(B+C))=cosAcosCsin(B+C)=cosAcosC

Hence, option (B) is correct.


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