If in a ∆ABC, acosA=bcosB, then
sin2A+sin2B=sin2C
2sinAcosB=sinC
2sinAsinBsinC=1
None of the above.
Explanation for the correct option:
Find the relation:
Given,
acosA=bcosB
Concept: We know the cosine rule,
a2=b2+c2–2bccosA,b2=a2+c2–2accosB,c2=a2+b2–2abcosC
acosA=bcosB⇒acosA=bcosB⇒a(a2+c2-b2)2ac=b(b2+c2-a2)2bc⇒a2+c2-b2=b2+c2-a2⇒2a2=2b2⇒a2=b2
Now, find the value of 2sinAcosB.
2sinAcosB=2×ak×(a2+c2-b2)2ac[∵asinA=bsinB=csinC=k]=2×ak×(b2+c2-b2)2ac[a2=b2,aboveproved]=c2ck=ck=sinC[∵asinA=bsinB=csinC=k]
Hence, the correct option is B.