If alpha and beta are solutions of sin^2 x + | vedclass

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

Question

(D) Given that $\alpha$ and $\beta$ are roots of $\sin^2 x + a \sin x + b = 0$,we have $\sin \alpha + \sin \beta = -a$.Given that $\alpha$ and $\beta$

Vedclass · Accepted Answer

$\frac{2ac}{a^2 + c^2}$

Vedclass · Answer

(D) Given that $\alpha$ and $\beta$ are roots of $\sin^2 x + a \sin x + b = 0$,we have $\sin \alpha + \sin \beta = -a$.Given that $\alpha$ and $\beta$ are roots of $\cos^2 x + c \cos x + d = 0$,we have $\cos \alpha + \cos \beta = -c$.Using sum-to-product formulas:$2 \sin \left( \frac{\alpha + \beta}{2} ight) \cos \left( \frac{\alpha - \beta}{2} ight) = -a$  $(1)$$2 \cos \left( \frac{\alpha + \beta}{2} ight) \cos \left( \frac{\alpha - \beta}{2} ight) = -c$  $(2)$Dividing $(1)$ by $(2)$:$	an \left( \frac{\alpha + \beta}{2} ight) = \frac{a}{c}$Using the identity $\sin 	heta = \frac{2 	an(	heta/2)}{1 + 	an^2(	heta/2)}$:$\sin(\alpha + \beta) = \frac{2 	an \left( \frac{\alpha + \beta}{2} ight)}{1 + 	an^2 \left( \frac{\alpha + \beta}{2} ight)} = \frac{2(a/c)}{1 + (a/c)^2} = \frac{2ac}{a^2 + c^2}$.

If $\alpha$ and $\beta$ are solutions of $\sin^2 x + a \sin x + b = 0$ as well as $\cos^2 x + c \cos x + d = 0$,then $\sin(\alpha + \beta)$ is equal to

Similar Questions

The set of solutions of the system of equations $x+y = \frac{2 \pi}{3}$ and $\cos x + \cos y = \frac{3}{2}$,where $x, y$ are real,is

If $A, B, C$ are the angles of a triangle,then $\sin^2 A + \sin^2 B + \sin^2 C - 2\cos A \cos B \cos C = $

In $\triangle PQR$,$\angle R = \frac{\pi}{4}$. If $\tan \left(\frac{P}{3}\right)$ and $\tan \left(\frac{Q}{3}\right)$ are the roots of the equation $ax^2 + bx + c = 0$,then:

If in triangle $ABC$,$\frac{a^2 - b^2}{a^2 + b^2} = \frac{\sin(A - B)}{\sin(A + B)}$,then the triangle is

The two adjacent sides of a cyclic quadrilateral are $2$ and $5$ and the angle between them is $60^{\circ}$. If the area of the quadrilateral is $4\sqrt{3}$,then the perimeter of the quadrilateral is

Vedclass Test Series

Exam Paper Generator

Online Exam Module