Given that tan alpha and tan beta are the roo | vedclass

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

Question

(A) From the given quadratic equation $x^2 - px + q = 0$,we have the sum and product of roots as:$	an \alpha + 	an \beta = p$ $(i)$$	an \alpha 	an

Vedclass · Accepted Answer

$\frac{p^2}{p^2 + (1 - q)^2}$

Vedclass · Answer

(A) From the given quadratic equation $x^2 - px + q = 0$,we have the sum and product of roots as:$	an \alpha + 	an \beta = p$ $(i)$$	an \alpha 	an \beta = q$ $(ii)$Using the formula for $	an(\alpha + \beta)$:$	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta} = \frac{p}{1 - q}$We know that $\sin^2 	heta = \frac{	an^2 	heta}{1 + 	an^2 	heta}$.Substituting $	heta = \alpha + \beta$:$\sin^2(\alpha + \beta) = \frac{	an^2(\alpha + \beta)}{1 + 	an^2(\alpha + \beta)}$$= \frac{(\frac{p}{1 - q})^2}{1 + (\frac{p}{1 - q})^2}$$= \frac{\frac{p^2}{(1 - q)^2}}{\frac{(1 - q)^2 + p^2}{(1 - q)^2}}$$= \frac{p^2}{p^2 + (1 - q)^2}$.

Given that $\tan \alpha$ and $\tan \beta$ are the roots of $x^2 - px + q = 0$,then the value of $\sin^2(\alpha + \beta)$ is:

Similar Questions

Two students were solving a quadratic equation in $x$. One student copied the constant term incorrectly and obtained the roots $3$ and $2$. The other student copied the constant term and the coefficient of $x^2$ correctly as $-6$ and $1$ respectively. What are the correct roots?

If the roots of the given equation $(2k + 1)x^2 - (7k + 3)x + k + 2 = 0$ are reciprocal to each other,then the value of $k$ will be

If $\alpha$ and $\beta$ are the roots of the equation $2x(2x+1)=1$,then $\beta$ is equal to

If the product of the roots of $9x^3 + 112x^2 - 120x + a = 0$ is $12$,then the value of '$a$' is

If $\alpha, \beta$ are the roots of the equation $x^2 - px + r = 0$ and $\alpha/2, 2\beta$ are the roots of the equation $x^2 - qx + r = 0$,then what is the value of $r$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module