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

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

Question

According to the given condition. $\sin \alpha+\sin \beta=-a$ and $\cos \alpha+\cos \beta=-c$

$\Rightarrow 2 \sin \frac{\alpha+\beta}{2} \cos \frac

Vedclass · Accepted Answer

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

Vedclass · Answer

According to the given condition. $\sin \alpha+\sin \beta=-a$ and $\cos \alpha+\cos \beta=-c$

$\Rightarrow 2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=-\mathrm{a}$      .....$(1)$

$	ext { and } 2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=-\mathrm{c}$        .....$(2)$

$(1) \div(2)$

$\Rightarrow 	an \frac{\alpha+\beta}{2}=\frac{a}{c}$

$\Rightarrow \sin (\alpha+\beta)=\frac{2 	an \frac{\alpha+\beta}{2}}{1+	an ^{2} \frac{\alpha+\beta}{2}}=\frac{2 a c}{a^{2}+c^{2}}$

Hence, $( 4)$ is the correct answer.

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

Similar Questions

$\frac{{\sec \,8\theta - 1}}{{\sec \,4\theta - 1}}$ is equal to

Prove that $\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}=\cot 3 x$

$\cos \frac{\pi }{7}\cos \frac{{2\pi }}{7}\cos \frac{{4\pi }}{7} = $

Prove that $=\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x$

The value of $\cos \left(\frac{2 \pi}{7}\right)+\cos \left(\frac{4 \pi}{7}\right)+\cos \left(\frac{6 \pi}{7}\right)$ is equal to