If cos (x-y), cos x, cos (x+y) are three dist | vedclass

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(A) Given that $\cos (x-y), \cos x, \cos (x+y)$ are in harmonic progression $(HP)$.Therefore,$\cos x = \frac{2 \cos (x-y) \cos (x+y)}{\cos (x+y) + \co

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$\cos ^2 x$

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(A) Given that $\cos (x-y), \cos x, \cos (x+y)$ are in harmonic progression $(HP)$.Therefore,$\cos x = \frac{2 \cos (x-y) \cos (x+y)}{\cos (x+y) + \cos (x-y)}$.Using the identity $\cos (A-B) \cos (A+B) = \cos ^2 A - \sin ^2 B = \cos ^2 A - (1 - \cos ^2 B) = \cos ^2 A + \cos ^2 B - 1$,we have:$\cos x = \frac{2(\cos ^2 x + \cos ^2 y - 1)}{2 \cos x \cos y}$.$\cos ^2 x \cos y = \cos ^2 x + \cos ^2 y - 1$.Rearranging the terms:$\cos ^2 x \cos y - \cos ^2 x = \cos ^2 y - 1$.$\cos ^2 x (\cos y - 1) = - (1 - \cos ^2 y)$.$\cos ^2 x (\cos y - 1) = - \sin ^2 y$.Since $1 - \cos ^2 y = \sin ^2 y$,we have:$\cos ^2 x (1 - \cos y) = (1 - \cos y)(1 + \cos y)$.Given $\cos x 
eq \cos y$,we can assume $\cos y 
eq 1$ (otherwise $\cos x = 1$,which contradicts the distinct numbers condition).Dividing both sides by $(1 - \cos y)$:$\cos ^2 x = 1 + \cos y$.

If $\cos (x-y), \cos x, \cos (x+y)$ are three distinct numbers which are in harmonic progression and $\cos x \neq \cos y$,then $1+\cos y$ is equal to

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