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