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

$$\begin{gathered} \mathrm{a} \sec \mathrm{\alpha }+\mathrm{b} \tan \mathrm{\alpha }=1 \\ {\mathrm{a}}^2 {\sec }^2\mathrm{\alpha }-{\mathrm{b}}^2 {\mathrm{tab}}^2\mathrm{\alpha }=5 \\ \mathrm{a} \sec \mathrm{\alpha }-\mathrm{b} \tan \mathrm{\alpha }=5 \\ 2\mathrm{a} \sec \mathrm{\alpha }=6 \Rightarrow \mathrm{a} \sec \mathrm{\alpha }=3 \\ 2\mathrm{b} \tan \mathrm{x}=-4 \Rightarrow \mathrm{b} \tan \mathrm{\alpha }=-2 \\ {\sec }^2\mathrm{\alpha }-{\tan }^2\mathrm{\alpha }=1 \\ \frac{9}{{\mathrm{a}}^2}-\frac{4}{{\mathrm{b}}^2}=1 \\ 9{\mathrm{b}}^2-4{\mathrm{a}}^2={\mathrm{a}}^2{\mathrm{b}}^2 \\ 4{\mathrm{a}}^2-9{\mathrm{b}}^2=-{\mathrm{a}}^2{\mathrm{b}}^2 \end{gathered}$$