If cos left(frac{alpha-beta}{2}right)=2 cos l | vedclass

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

Question

(C) Given,$\cos \left(\frac{\alpha-\beta}{2}\right)=2 \cos \left(\frac{\alpha+\beta}{2}\right)$.Applying the componendo and dividendo rule:$\frac{\cos

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(C) Given,$\cos \left(\frac{\alpha-\beta}{2}ight)=2 \cos \left(\frac{\alpha+\beta}{2}ight)$.Applying the componendo and dividendo rule:$\frac{\cos \left(\frac{\alpha-\beta}{2}ight) + \cos \left(\frac{\alpha+\beta}{2}ight)}{\cos \left(\frac{\alpha-\beta}{2}ight) - \cos \left(\frac{\alpha+\beta}{2}ight)} = \frac{2+1}{2-1}$.Using the sum-to-product formulas $\cos(A-B) + \cos(A+B) = 2 \cos A \cos B$ and $\cos(A-B) - \cos(A+B) = 2 \sin A \sin B$:$\frac{2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2}}{2 \sin \frac{\alpha}{2} \sin \frac{\beta}{2}} = \frac{3}{1}$.$\cot \frac{\alpha}{2} \cot \frac{\beta}{2} = 3$.Therefore,$	an \frac{\alpha}{2} 	an \frac{\beta}{2} = \frac{1}{3}$.

If $\cos \left(\frac{\alpha-\beta}{2}\right)=2 \cos \left(\frac{\alpha+\beta}{2}\right)$,then $\tan \frac{\alpha}{2} \tan \frac{\beta}{2}=$

Similar Questions

$\cos ^4 \frac{\pi}{12} + \cos ^4 \frac{5 \pi}{12} + \cos ^4 \frac{7 \pi}{12} + \cos ^4 \frac{11 \pi}{12} = $

The equation $\sin x \cos x = 2$ has

The variable $x$ satisfying the equation $|\sin x \cos x| + \sqrt{2 + \tan^2 x + \cot^2 x} = \sqrt{3}$ belongs to the interval

If $x$ and $y$ are acute angles such that $\cos x + \cos y = \frac{3}{2}$ and $\sin x + \sin y = \frac{3}{4}$,then $\sin(x + y)$ equals to

Let $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$,where $x \in R$ and $k \ge 1$. Then $f_4(x) - f_6(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module