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(D) Given that $	an \alpha$ and $	an \beta$ are the roots of $x^2 + px + q = 0$.By Vieta's formulas,$	an \alpha + 	an \beta = -p$ and $	an \alpha

Vedclass · Accepted Answer

Both $(a)$ and $(b)$

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(D) Given that $	an \alpha$ and $	an \beta$ are the roots of $x^2 + px + q = 0$.By Vieta's formulas,$	an \alpha + 	an \beta = -p$ and $	an \alpha 	an \beta = q$.Using the identity for $	an(\alpha + \beta)$:$	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta} = \frac{-p}{1 - q} = \frac{p}{q - 1}$.This confirms option $(b)$ is correct.Now,consider the expression in $(a)$:$E = \sin^2(\alpha + \beta) + p\sin(\alpha + \beta)\cos(\alpha + \beta) + q\cos^2(\alpha + \beta)$.Divide by $\cos^2(\alpha + \beta)$ and multiply by $\cos^2(\alpha + \beta)$:$E = \cos^2(\alpha + \beta) [	an^2(\alpha + \beta) + p	an(\alpha + \beta) + q]$.Since $\cos^2(\alpha + \beta) = \frac{1}{1 + 	an^2(\alpha + \beta)}$,we substitute $	an(\alpha + \beta) = \frac{p}{q - 1}$:$E = \frac{1}{1 + (\frac{p}{q - 1})^2} [(\frac{p}{q - 1})^2 + p(\frac{p}{q - 1}) + q]$$E = \frac{(q - 1)^2}{(q - 1)^2 + p^2} [\frac{p^2 + p^2(q - 1) + q(q - 1)^2}{(q - 1)^2}]$$E = \frac{p^2 + p^2q - p^2 + q(q^2 - 2q + 1)}{p^2 + (q - 1)^2} = \frac{p^2q + q^3 - 2q^2 + q}{p^2 + (q - 1)^2} = \frac{q(p^2 + q^2 - 2q + 1)}{p^2 + (q - 1)^2} = \frac{q(p^2 + (q - 1)^2)}{p^2 + (q - 1)^2} = q$.Thus,$(a)$ is also correct. Therefore,the correct option is $(d)$.

If $\tan \alpha$ and $\tan \beta$ are the roots of the equation $x^2 + px + q = 0$ $(p \neq 0)$,then:

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