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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]

Refer to the attachment

View image Аноним
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