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

If, $\beta =\underset{x\to 0}{\lim }{\left(\cos x\right)}^{\cot x}$    ($1^{\infty }$form)

$\beta =e^{ \underset{x\to 0}{\lim }\cot x\left[\cos x-1\right]}$

$\beta =e^{ \underset{x\to 0}{\lim }\frac{-\cos x}{\sin x}\left[1-1+2{\sin }^2\frac{x}{2}\right]}$

$\beta =e^{ \underset{x\to 0}{\lim } \frac{-\cos x}{2\sin \frac{x}{2}.\cos \frac{x}{2}}\left[1-1+2{\sin }^2\frac{x}{2}\right]}$

$\beta =e^{ \underset{x\to 0}{\lim } \frac{-\cos x.\sin {\displaystyle \frac{x}{2}}}{\cos \frac{x}{2}}}$

$\beta =e^0$

$\beta =1$
$\alpha =\underset{x\to \pi /4}{\lim }\frac{{\tan }^{3}x-\tan x}{\cos \left(x+\frac{\pi }{4}\right)}$

$=\underset{x\to \pi /4}{\lim }\frac{\tan x(\tan x+1)(\tan x-1)}{\cos \left(x+\frac{\pi }{4}\right)}$

$\alpha =\underset{x\to \pi /4}{\lim }\tan x\times \underset{x\to \pi /4}{\lim }\left(\tan x+1\right)\times \underset{x\to \pi /4}{\lim }\frac{(\tan x-1)}{\cos \left(x+{\displaystyle \frac{\mathrm{\pi }}{4}}\right)}$

$\alpha =1\times \left(1+1\right)\times \underset{x\to \pi /4}{\lim }\frac{(\tan x-1)}{\cos \left(x+{\displaystyle \frac{\mathrm{\pi }}{4}}\right)}$

$=2\underset{x\to \pi /4}{\lim }\frac{\tan x-1}{\cos \left(x+\frac{\pi }{4}\right)}$

$=2\underset{x\to \pi /4}{\lim }\frac{\sin x-\cos x}{\cos x.\cos \left(x+\frac{\pi }{4}\right)}$

$=-2\sqrt{2}\underset{x\to \pi /4}{\lim }\frac{\left({\displaystyle \frac{1}{\sqrt{2}}}\cos x-\frac{1}{\sqrt{2}}\sin x\right)}{\cos x.\cos \left(x+\frac{\pi }{4}\right)}$

$\alpha =-2\sqrt{2}\underset{x\to \pi /4}{\lim }\frac{\cos \left(x+\frac{\pi }{4}\right)}{\cos x.\cos \left(x+\frac{\pi }{4}\right)}$

$\alpha =-4$

So, the equation whose roots are $\alpha$ and $\beta$ is

$x^2+3x-4=0$$$

Hence, $a=1, b=3$