Answer:
Solutions given:
[tex] \frac{SinA}{SinB}=p[/tex]
SinA=pSin B or
Sin B=Sin A/p
[tex]\frac{CosA}{CosB}=q[/tex]
Cos A=qCos B
or
Cos B=Cos A /q
now
TanA=[tex]\frac{Sin A}{Cos A}=\frac{psinB}{qCosB}=\frac{p}{q}*TanB[/tex]
again
TanB=[tex]\frac{Sin B}{Cos B}=\frac{Sin A /p}{CosA/q}=\frac{q}{p} Tan A[/tex]
now
[tex]\frac{TanA}{TanB}=\frac{p}{q}[/tex]
