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 $2 \cos A \cos B = \cos (A+B) + \cos (A-B)$,we have:$\cos x = \frac{\cos 2x + \cos 2y}{2 \cos x \cos y}$.Using $\cos 2	heta = 2 \cos^2 	heta - 1$,we get:$\cos x = \frac{(2 \cos^2 x - 1) + (2 \cos^2 y - 1)}{2 \cos x \cos y} = \frac{2 \cos^2 x + 2 \cos^2 y - 2}{2 \cos x \cos y}$.$\cos x = \frac{\cos^2 x + \cos^2 y - 1}{\cos x \cos y}$.$\cos^2 x \cos y = \cos^2 x + \cos^2 y - 1$.$\cos^2 x \cos y - \cos^2 x = \cos^2 y - 1$.$-\cos^2 x (1 - \cos y) = -(1 - \cos^2 y)$.$\cos^2 x (1 - \cos y) = (1 - \cos y)(1 + \cos y)$.Since $\cos x 
eq \cos y$,we have $1 - \cos y 
eq 0$,so we can divide 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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