Question

If $\frac{\sin A}{\sin B}=p$ and $\frac{\cos A}{\cos B}=q$, find $\tan A$ and $\tan B$.

Answer 1

$\frac{sinA}{sinB}=p⟹sinA=psinB$ ----(1)

$\frac{cosA}{cosB}=q⟹cosA=qcosB$-----(2)

$si{n}^{2}A+co{s}^{2}A=1$

From (1) and (2), $p^{2}si{n}^{2}B+q^{2}co{s}^{2}B=1$

Also $si{n}^{2}B+co{s}^{2}B=1$

Equating, $p^{2}si{n}^{2}B+q^{2}co{s}^{2}B=si{n}^{2}B+co{s}^{2}B$

Divide by $co{s}^{2}B$, $p^{2}ta{n}^{2}B+q^2=ta{n}^{2}B+1$

$ta{n}^{2}B=\frac{1-q^2}{p^2-1}$

$⟹tanB=\sqrt{\frac{1-q^2}{p^2-1}}$ ---(3)

Dividing (1) by (2) $tanA=\frac{p}{q}tanB$

Substituting $tanB$ from (3),

$tanA=\frac{p}{q}\sqrt{\frac{1-q^2}{p^2-1}}$