Let alpha and beta be nonzero real numbers su | vedclass

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(A,C) Given $2(\cos \beta - \cos \alpha) + \cos \alpha \cos \beta = 1$.Rearranging the terms,we get $\cos \beta(2 + \cos \alpha) = 1 + 2 \cos \alpha$.

Vedclass · Accepted Answer

$	an \left(\frac{\alpha}{2}ight) + \sqrt{3} 	an \left(\frac{\beta}{2}ight) = 0$

Vedclass · Answer

(A,C) Given $2(\cos \beta - \cos \alpha) + \cos \alpha \cos \beta = 1$.Rearranging the terms,we get $\cos \beta(2 + \cos \alpha) = 1 + 2 \cos \alpha$.So,$\cos \beta = \frac{1 + 2 \cos \alpha}{2 + \cos \alpha}$.Applying componendo and dividendo,$\frac{\cos \beta - 1}{\cos \beta + 1} = \frac{(1 + 2 \cos \alpha) - (2 + \cos \alpha)}{(1 + 2 \cos \alpha) + (2 + \cos \alpha)} = \frac{\cos \alpha - 1}{3(1 + \cos \alpha)}$.Using the identities $\cos 	heta - 1 = -2 \sin^2(	heta/2)$ and $\cos 	heta + 1 = 2 \cos^2(	heta/2)$,we get $\frac{-2 \sin^2(\beta/2)}{2 \cos^2(\beta/2)} = \frac{-2 \sin^2(\alpha/2)}{3(2 \cos^2(\alpha/2))}$.This simplifies to $-	an^2(\beta/2) = -\frac{1}{3} 	an^2(\alpha/2)$,which implies $	an^2(\alpha/2) = 3 	an^2(\beta/2)$.Taking the square root,$	an(\alpha/2) = \pm \sqrt{3} 	an(\beta/2)$.Thus,$	an(\alpha/2) - \sqrt{3} 	an(\beta/2) = 0$ or $	an(\alpha/2) + \sqrt{3} 	an(\beta/2) = 0$.

Let $\alpha$ and $\beta$ be nonzero real numbers such that $2(\cos \beta - \cos \alpha) + \cos \alpha \cos \beta = 1$. Then which of the following is/are true?

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