Question

If $\frac{\sin A}{\sin B}=p,\frac{\cos A}{\cos B}=q,$ then prove that -
$\tan A=\pm \frac{p}{q}\sqrt{\left(\frac{1-q^2}{p^2-1}\right)},\tan B=\pm \sqrt{\left(\frac{1-q^2}{p^2-1}\right)}$

Answer 1

let $\tan =A$ $t_1,\tan B=t_2$
Dividing the given relation
$\frac{\tan A}{\tan B}=\frac{p}{q}or\frac{t_1}{p}=\frac{t_2}{q}=K$ , say ....(1)
multiplying the given relation, $\frac{\sin 2A}{\sin 2B}=pq$
or $\frac{2t_1}{1+t_1^2}\cdot \frac{1+t_2^2}{2t_2}=pq$
Now put for $t_1$ and $t_2$ from (1) and find K and hence
$t_1=\pm \frac{p}{q}\sqrt{\left(\frac{1-q^2}{p^2-1}\right)},t_2=\pm \sqrt{\left(\frac{1-q^2}{p^2-1}\right)}$