If $\text{asec}\alpha \text{ + btan}\alpha \text{ =1}$ and $a^2{\sec }^2\alpha -b^2{\tan }^2\alpha =5$  then $\text{4}{a}^2–\text{9}{b}^2 =$

$\left(a\sec \alpha +b\tan \alpha \right)\left(a\sec \alpha -b\tan \alpha \right)=5$ 

            $\Rightarrow a\sec \alpha -b\tan \alpha =5$

            $\Rightarrow -b\tan \alpha -b\tan \alpha =5-1$

            $\Rightarrow \tan \alpha =\frac{-2}{b}$

            $\therefore a\sec \alpha +b\left(\frac{-2}{b}\right)=1$

            $\Rightarrow \sec \alpha =\frac{3}{a}$

            ${\sec }^2\alpha -{\tan }^2\alpha =1$

            $\Rightarrow \frac{9}{a^2}-\frac{4}{b^2}=1$

                $\Rightarrow 4a^2-9b^2=-a^2{b}^2$