If $a{\sin }^2\theta +b{\cos }^2\theta =a{\cos }^2\varphi +b{\sin }^2\varphi =1$ 

and $a\tan \theta =b\tan \varphi (a\ne b)$ then 

$a{\tan }^2\theta +b=1+{\tan }^2\theta \Rightarrow (a-1){\tan }^2\theta =1-b$ 

and $(b-1){\tan }^2\varphi =1-a\Rightarrow \frac{{\tan }^2\theta }{{\tan }^2\varphi }=\frac{(1-b{)}^2}{(1-a{)}^2}=\frac{b^2}{a^2}$ 

$\begin{array}{r}\Rightarrow (a-b)(a+b-2ab)=0\\ \Rightarrow a+b=2ab\text{ [as }a\ne b]\end{array}$