If $\sin \alpha =a\sin \beta and\cos \alpha =b\cos \beta then\tan \alpha =$

$\begin{array}{l}1={\sin }^2\beta +{\cos }^2\beta =\frac{{\sin }^2\alpha }{a^2}+\frac{{\cos }^2\alpha }{b^2}\\ \Rightarrow {\sec }^2\alpha =\frac{{\tan }^2\alpha }{a^2}+\frac{1}{b^2}\Rightarrow a^2{b}^2(1+{\tan }^2\alpha )=b^2{\tan }^2\alpha +a^2\\ \Rightarrow (b^2-a^2{b}^2){\tan }^2\alpha =a^2{b}^2-a^2\\ \Rightarrow {\tan }^2\alpha =\frac{a^2{b}^2-a^2}{b^2-a^2{b}^2}\Rightarrow \tan \alpha =\pm \sqrt{\left(\frac{a^2{b}^2-a^2}{b^2-a^2{b}^2}\right)}.\end{array}$