The quadratic equation whose roots are l and | vedclass

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(B) First,we calculate $l$: $l = \lim_{	heta ightarrow 0} \frac{3 \sin 	heta - 4 \sin^2 	heta}{	heta} = \lim_{	heta ightarrow 0} \left( 3 \fr

Vedclass · Accepted Answer

$x^2 - 5x + 6 = 0$

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(B) First,we calculate $l$: $l = \lim_{	heta ightarrow 0} \frac{3 \sin 	heta - 4 \sin^2 	heta}{	heta} = \lim_{	heta ightarrow 0} \left( 3 \frac{\sin 	heta}{	heta} - 4 \sin 	heta \cdot \frac{\sin 	heta}{	heta} ight) = 3(1) - 4(0)(1) = 3$. Next,we calculate $m$: $m = \lim_{	heta ightarrow 0} \frac{2 	an 	heta}{	heta(1 - 	an^2 	heta)} = \lim_{	heta ightarrow 0} \left( \frac{	an 	heta}{	heta} \cdot \frac{2}{1 - 	an^2 	heta} ight) = 1 \cdot \frac{2}{1 - 0} = 2$. The quadratic equation with roots $l=3$ and $m=2$ is given by $x^2 - (l+m)x + lm = 0$. Substituting the values,we get $x^2 - (3+2)x + (3 	imes 2) = 0$,which simplifies to $x^2 - 5x + 6 = 0$.

The quadratic equation whose roots are $l$ and $m$,where $l = \lim_{\theta \rightarrow 0} \left( \frac{3 \sin \theta - 4 \sin^2 \theta}{\theta} \right)$ and $m = \lim_{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta(1 - \tan^2 \theta)}$,is:

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