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If a/SinA = b/CosA then prove that SinACosA = ab/a2 + b2
Dear sunny
we know by sine rule that
a/sinA=b/sinB=c/sinC
futher it is given that
a/sinA=b/cosA
hence b/sinB=b/cosA
So cosA=sinB
cosA=cos(90-B)
hence A+B=90
So C=90
Suppose
a/sinA=b/sinB=c/sinC=k
sinA=a/k;sinB=b/k;sinC=c/k
so sinA sinB/sin2 C=ab/c2
but we know Sin B =Cos A and c2=a2 +b2 and SinC=1 (since triangle is right angled at C)
so sin A cosA=ab/a2 + b2
ALL THE BEST
Abhinav Batra
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