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(A) Given the quadratic equation $6x^2 - 5x + 1 = 0$.For a quadratic equation $ax^2 + bx + c = 0$,the sum of roots $\alpha + \beta = -b/a$ and the pro

Vedclass · Accepted Answer

$\pi / 4$

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(A) Given the quadratic equation $6x^2 - 5x + 1 = 0$.For a quadratic equation $ax^2 + bx + c = 0$,the sum of roots $\alpha + \beta = -b/a$ and the product of roots $\alpha \beta = c/a$.Here,$\alpha + \beta = -(-5)/6 = 5/6$ and $\alpha \beta = 1/6$.We use the identity $	an^{-1} \alpha + 	an^{-1} \beta = 	an^{-1} \left( \frac{\alpha + \beta}{1 - \alpha \beta} ight)$.Substituting the values: $	an^{-1} \left( \frac{5/6}{1 - 1/6} ight) = 	an^{-1} \left( \frac{5/6}{5/6} ight) = 	an^{-1}(1)$.Since $	an^{-1}(1) = \pi / 4$,the final value is $\pi / 4$.

If $\alpha$ and $\beta$ are the roots of the equation $6x^2 - 5x + 1 = 0$,then the value of $\tan^{-1} \alpha + \tan^{-1} \beta$ is:

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