Question

If $\frac{\sin A}{\sin B}=p$ and $\frac{\cos A}{\cos B}=q$, then $tanA$ and $tanB$ .

Answer 1

$\frac{\sin A}{\sin B}=p;\frac{\cos A}{\cos B}=q$

$\sin A=p\sin B---\left(1\right);\cos A=q\cos B---\left(2\right)$

$\tan A=\frac{p}{q}\tan B---\left(3\right)\left(dividing\left(1\right)by\left(2\right)\right)$$\sin A\cos A=pq\sin B\cos B\left(Multiply\left(1\right)\&\left(2\right)\right)$

$\frac{\sin A\cos A}{{\cos }^{2}A{\cos }^{2}B}=\frac{pq\sin B\cos B}{{\cos }^{2}A{\cos }^{2}B}$

$={\sec }^{2}B\tan A=pq{\sec }^{2}A\tan B$

$=\left(1+{\tan }^{2}B\right)\tan A=pq\left(1+{\tan }^{2}A\right)\tan B$

$=\left[1+{\left(\frac{p}{q}\tan A\right)}^2\right]\tan A=pq\left(1+{\tan }^{2}A\right).\frac{p}{q}\tan A\left(by\left(3\right)\right)$

$=1+\frac{p^2}{q^2}{\tan }^{2}A=q^2+q^2{\tan }^{2}A$

$={\tan }^{2}A\left(\frac{q^2}{p^2}-q^2\right)=q^2-1$

$={\tan }^{2}A\frac{\left(q^2-1\right)p^2}{q^2-p^2{q}^2}$

$\tan A=\sqrt{\frac{p^2\left(1-q^2\right)}{q^2\left(p^2-1\right)}}=\frac{p}{q}\sqrt{\frac{1-q^2}{p^2-1}}$

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