If tan alpha and tan beta are the roots of th | vedclass

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(D) Given that $	an \alpha$ and $	an \beta$ are the roots of the equation $x^2 + px + q = 0$.By the properties of roots,we have:$	an \alpha + 	an

Vedclass · Accepted Answer

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

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(D) Given that $	an \alpha$ and $	an \beta$ are the roots of the equation $x^2 + px + q = 0$.By the properties of roots,we have:$	an \alpha + 	an \beta = -p$ and $	an \alpha 	an \beta = q$.Using the tangent addition formula:$	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta} = \frac{-p}{1 - q} = \frac{p}{q - 1}$.This confirms that option $(b)$ is correct.Now,consider the expression in $(a)$:$L.H.S. = \cos^2(\alpha + \beta) [	an^2(\alpha + \beta) + p	an(\alpha + \beta) + q]$.Since $\cos^2(\alpha + \beta) = \frac{1}{1 + 	an^2(\alpha + \beta)}$,we substitute $	an(\alpha + \beta) = \frac{p}{q - 1}$:$L.H.S. = \frac{1}{1 + \frac{p^2}{(q - 1)^2}} \left[ \frac{p^2}{(q - 1)^2} + p\left(\frac{p}{q - 1}ight) + q ight]$.Simplifying the expression inside the bracket:$= \frac{(q - 1)^2}{(q - 1)^2 + p^2} \left[ \frac{p^2 + p^2(q - 1) + q(q - 1)^2}{(q - 1)^2} ight] = \frac{p^2 + p^2q - p^2 + q(q^2 - 2q + 1)}{(q - 1)^2 + p^2} = \frac{p^2q + q^3 - 2q^2 + q}{(q - 1)^2 + p^2}$.Since $	an \alpha, 	an \beta$ are roots of $x^2 + px + q = 0$,the discriminant $D = p^2 - 4q \ge 0$,so $p^2 \ge 4q$. Substituting $p^2 = -q - 	an^2...$ is not needed; simply note that the expression evaluates to $q$.Thus,both $(a)$ and $(b)$ are correct.

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

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