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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $(b)$

Vedclass · Answer

(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:

Similar Questions

The minimum value of $9\tan^2\theta + 4\cot^2\theta$ is

If $\cos \theta + \sin \theta = \sqrt{2} \cos \theta$,then $\cos \theta - \sin \theta =$

The value of $\sec ^{2} 12^{\circ}-\frac{1}{\tan ^{2} 78^{\circ}}$ is

If $\tan \alpha = \frac{m}{m+1}$ and $\tan \beta = \frac{1}{2m+1}$,then $\alpha + \beta =$

$\sin ^2 \frac{\pi }{8} + \sin ^2 \frac{3\pi }{8} + \sin ^2 \frac{5\pi }{8} + \sin ^2 \frac{7\pi }{8} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module