If $\mathrm{\alpha }=\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{4}}{lim} \frac{{\tan }^3\mathrm{x}-\tan \mathrm{x}}{\cos \left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)}$ and $\mathrm{\beta }=\underset{\mathrm{x}\to 0}{lim} (\cos \mathrm{x}{)}^{\cot \mathrm{x}}$ are the roots of the equation, ${\mathrm{ax}}^2+\mathrm{bx}-4=0$, then the ordered pair (a, b) is

$\begin{array}{l}\mathrm{\alpha }=\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{4}}{lim} \frac{{\tan }^3\mathrm{x}-\tan \mathrm{x}}{\cos \left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)}\\ =\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{4}}{lim} \frac{\tan \mathrm{x}(\tan \mathrm{x}+1)(\tan \mathrm{x}-1)}{\cos \left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)}\\ =\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{4}}{lim} \frac{\frac{\sin \mathrm{x}}{\cos \mathrm{x}}\cdot \left(\frac{\sin \mathrm{x}-\cos \mathrm{x}}{\cos \mathrm{x}}\right)\left(\frac{\sin \mathrm{x}+\cos \mathrm{x}}{\cos \mathrm{x}}\right)}{\frac{1}{\sqrt{2}}(\cos \mathrm{x}-\sin \mathrm{x})}\\ =\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{4}}{lim} \frac{-\sqrt{2}\sin \mathrm{x}(\sin \mathrm{x}+\cos \mathrm{x})}{{\cos }^3\mathrm{x}}\\ =\frac{-\sqrt{2}\times \frac{1}{\sqrt{2}}\times \sqrt{2}}{\frac{1}{2\sqrt{2}}}=-4\end{array}$

$\begin{array}{l}\mathrm{and} \mathrm{\beta }=\underset{\mathrm{x}\to 0}{lim} (\cos \mathrm{x}{)}^{\cot \mathrm{x}}\\ ={\mathrm{e}}^{\underset{\mathrm{x}\to 0}{lim} \frac{\cos \mathrm{x}-1}{\tan \mathrm{x}}}={\mathrm{e}}^{\underset{\mathrm{x}\to 0}{lim} -\frac{\sin \mathrm{x}}{{\sec }^2\mathrm{x}}}={\mathrm{e}}^0=1\end{array}$

Equation whose roots are α and β, is

$\begin{array}{l} {\mathrm{x}}^2+3\mathrm{x}-4=0\\ \therefore \mathrm{a}=1, \mathrm{b}=3\end{array}$