Let alpha and beta be the roots of the quadra | vedclass

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

Question

(C) The given quadratic equation is $x^2 \sin 	heta - x(\sin 	heta \cos 	heta + 1) + \cos 	heta = 0$.Using the quadratic formula $x = \frac{-b \pm

Vedclass · Accepted Answer

$\frac{1}{1 - \cos 	heta} + \frac{1}{1 + \sin 	heta}$

Vedclass · Answer

(C) The given quadratic equation is $x^2 \sin 	heta - x(\sin 	heta \cos 	heta + 1) + \cos 	heta = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we have:$x = \frac{(\sin 	heta \cos 	heta + 1) \pm \sqrt{(\sin 	heta \cos 	heta + 1)^2 - 4 \sin 	heta \cos 	heta}}{2 \sin 	heta}$Since $(\sin 	heta \cos 	heta + 1)^2 - 4 \sin 	heta \cos 	heta = (\sin 	heta \cos 	heta - 1)^2$,we get:$x = \frac{(\sin 	heta \cos 	heta + 1) \pm (\sin 	heta \cos 	heta - 1)}{2 \sin 	heta}$Taking the positive sign: $x = \frac{2 \sin 	heta \cos 	heta}{2 \sin 	heta} = \cos 	heta$.Taking the negative sign: $x = \frac{2}{2 \sin 	heta} = \csc 	heta$.Given $0  \sin 	heta$,so $\cos 	heta The sum is $\sum_{n=0}^\infty \alpha^n + \sum_{n=0}^\infty (-\frac{1}{\beta})^n = \sum_{n=0}^\infty (\cos 	heta)^n + \sum_{n=0}^\infty (-\sin 	heta)^n$.Using the infinite geometric series formula $\sum_{n=0}^\infty r^n = \frac{1}{1-r}$:Sum $= \frac{1}{1 - \cos 	heta} + \frac{1}{1 - (-\sin 	heta)} = \frac{1}{1 - \cos 	heta} + \frac{1}{1 + \sin 	heta}$.

Let $\alpha$ and $\beta$ be the roots of the quadratic equation $x^2 \sin \theta - x(\sin \theta \cos \theta + 1) + \cos \theta = 0$ where $0 < \theta < 45^\circ$,and $\alpha < \beta$. Then $\sum_{n=0}^\infty (\alpha^n + \frac{(-1)^n}{\beta^n})$ is equal to

Similar Questions

Solve the given two equations and choose the correct option.
$I. 20 x^{2}-67 x+56=0$
$II. 56 y^{2}-67 y+20=0$

Let $\alpha, \alpha^2$ be the roots of $x^2 + x + 1 = 0$. Then the equation whose roots are $\alpha^{31}, \alpha^{62}$ is:

Find the value of $k$ so that the sum of the roots of the equation $3 x^{2}+(2 k+1) x-k-5=0$ is equal to the product of the roots:

The existence of a unique solution for the system of equations $x+y+z=\beta$,$5x-y+\alpha z=10$,and $2x+3y-z=6$ depends on:

If the ratio of the roots of $ax^2 + 2bx + c = 0$ is the same as the ratio of the roots of $px^2 + 2qx + r = 0$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module