If y = frac{x^4 - x^2 + 1}{x^2 + sqrt{3}x + 1 | vedclass

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(B) Given $y = \frac{x^4 - x^2 + 1}{x^2 + \sqrt{3}x + 1}$.We can rewrite the numerator as $x^4 + 2x^2 + 1 - 3x^2 = (x^2 + 1)^2 - (\sqrt{3}x)^2$.Using

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$\cot\frac{5\pi}{12}$

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(B) Given $y = \frac{x^4 - x^2 + 1}{x^2 + \sqrt{3}x + 1}$.We can rewrite the numerator as $x^4 + 2x^2 + 1 - 3x^2 = (x^2 + 1)^2 - (\sqrt{3}x)^2$.Using the identity $a^2 - b^2 = (a - b)(a + b)$,we get $y = \frac{(x^2 + 1 - \sqrt{3}x)(x^2 + 1 + \sqrt{3}x)}{x^2 + \sqrt{3}x + 1}$.Simplifying the expression,we have $y = x^2 - \sqrt{3}x + 1$.Now,differentiating with respect to $x$,we get $\frac{dy}{dx} = 2x - \sqrt{3}$.Comparing this with $\frac{dy}{dx} = ax + b$,we find $a = 2$ and $b = -\sqrt{3}$.Thus,$a + b = 2 - \sqrt{3}$.Since $	an(15^\circ) = 	an(\frac{\pi}{12}) = 2 - \sqrt{3}$,and $	an(\frac{\pi}{12}) = \cot(\frac{\pi}{2} - \frac{\pi}{12}) = \cot(\frac{5\pi}{12})$.Therefore,$a + b = \cot\frac{5\pi}{12}$.

If $y = \frac{x^4 - x^2 + 1}{x^2 + \sqrt{3}x + 1}$ and $\frac{dy}{dx} = ax + b$,then the value of $a + b$ is equal to

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