If cos alpha = frac{2 cos beta - 1}{2 - cos b | vedclass

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

Question

(A) Given $\cos \alpha = \frac{2 \cos \beta - 1}{2 - \cos \beta}$.Applying Componendo and Dividendo:$\frac{1 - \cos \alpha}{1 + \cos \alpha} = \frac{(

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(A) Given $\cos \alpha = \frac{2 \cos \beta - 1}{2 - \cos \beta}$.Applying Componendo and Dividendo:$\frac{1 - \cos \alpha}{1 + \cos \alpha} = \frac{(2 - \cos \beta) - (2 \cos \beta - 1)}{(2 - \cos \beta) + (2 \cos \beta - 1)}$$\frac{2 \sin^2 \frac{\alpha}{2}}{2 \cos^2 \frac{\alpha}{2}} = \frac{3 - 3 \cos \beta}{1 + \cos \beta}$$	an^2 \frac{\alpha}{2} = \frac{3(1 - \cos \beta)}{1 + \cos \beta}$$	an^2 \frac{\alpha}{2} = 3 	an^2 \frac{\beta}{2}$$	an^2 \frac{\alpha}{2} \cot^2 \frac{\beta}{2} = 3$Taking the square root,$	an \frac{\alpha}{2} \cot \frac{\beta}{2} = \sqrt{3}$.

If $\cos \alpha = \frac{2 \cos \beta - 1}{2 - \cos \beta}$,then the value of $\tan \frac{\alpha}{2} \cot \frac{\beta}{2}$ is equal to,where $(0 < \alpha < \pi$ and $0 < \beta < \pi)$.

Similar Questions

The product $(1+\tan 1^{\circ})(1+\tan 2^{\circ})(1+\tan 3^{\circ}) \dots (1+\tan 45^{\circ})$ equals

The value of $\cos \frac{\pi}{10} \cos \frac{2\pi}{10} \cos \frac{4\pi}{10} \cos \frac{8\pi}{10} \cos \frac{16\pi}{10}$ is

The number of real roots of the equation $e^{\sin x} - e^{-\sin x} - 4 = 0$ is

If $\frac{\sin (A+B)}{\sin (A-B)}=\frac{\cos (C+D)}{\cos (C-D)}$,then $\tan A \cot B=$

In a triangle $ABC$,if $3 \sin A + 4 \cos B = 6$ and $4 \sin B + 3 \cos A = 1$,then $\sin (A + B)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module