If $$\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{{\tan }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$$ and $\beta = \mathop {\lim }\limits_{x \to 0 } {(\cos x)^{\cot x}}$ are the roots of the equation, ax² + bx $-$ 4 = 0, then the ordered pair (a, b) is :

$$\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{{\tan }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}};{0 \over 0}$$ form<br><br>Using L Hospital rule<br><br>$$\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{3{{\tan }^2}x{{\sec }^2}x - {{\sec }^2}x} \over { - \sin \left( {x + {\pi \over 4}} \right)}}$$<br><br>$\alpha$ = $-$4<br><br>$$\beta = \mathop {\lim }\limits_{x \to 0} {(\cos x)^{\cot x}} = {e^{\mathop {\lim }\limits_{x \to 0} {{(\cos x - 1)} \over {\tan x}}}}$$<br><br>$$\beta = {e^{\mathop {\lim }\limits_{x \to 0} {{ - (1 - \cos x)} \over {{x^2}}}.{{{x^2}} \over {{{\left( {{{\tan x} \over x}} \right)}^x}}}}}$$<br><br>$$\beta = {e^{\mathop {\lim }\limits_{x \to 0} \left( {{{ - 1} \over 2}} \right).{x \over 1}}} = {e^0} \Rightarrow \beta = 1$$<br><br>$\alpha$ = $-$4; $\beta$ = 1<br><br>If ax² + bx $-$ 4 = 0 are the roots then<br><br>16a $-$ 4b $-$ 4 = 0 &amp; a + b $-$ 4 = 0<br><br>$\Rightarrow$ a = 1 &amp; b = 3