If alpha and beta are the roots of the quadra | vedclass

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(B) Given the quadratic equation $3x^2 - 16x + 5 = 0$. Comparing with $ax^2 + bx + c = 0$,we have $a = 3, b = -16, c = 5$. Sum of roots $\alpha + \bet

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$\pi$

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(B) Given the quadratic equation $3x^2 - 16x + 5 = 0$. Comparing with $ax^2 + bx + c = 0$,we have $a = 3, b = -16, c = 5$. Sum of roots $\alpha + \beta = -\frac{b}{a} = -\left(\frac{-16}{3}ight) = \frac{16}{3}$. Product of roots $\alpha \beta = \frac{c}{a} = \frac{5}{3}$. Since $\alpha \beta = \frac{5}{3} > 1$,we use the formula: $	an^{-1} \alpha + 	an^{-1} \beta = \pi + 	an^{-1}\left(\frac{\alpha + \beta}{1 - \alpha \beta}ight)$. Substituting this into the given expression: $\left[\pi + 	an^{-1}\left(\frac{\alpha + \beta}{1 - \alpha \beta}ight)ight] - 	an^{-1}\left(\frac{\alpha + \beta}{1 - \alpha \beta}ight) = \pi$.

If $\alpha$ and $\beta$ are the roots of the quadratic equation $3x^2 - 16x + 5 = 0$,then $\tan^{-1} \alpha + \tan^{-1} \beta - \tan^{-1}\left(\frac{\alpha + \beta}{1 - \alpha \beta}\right) = $

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