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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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The correct answer is:
$\cos ^2 x$
$\cos (x-y), \cos x, \cos (x+y)$ are in HP.
Then, $\cos x=\frac{2 \cos (x-y) \cos (x+y)}{\cos (x+y)+\cos (x-y)}$
$\cos x=\frac{\cos 2 x+\cos 2 y}{2 \cos x \cdot \cos y}$
$\cos x=\frac{2 \cos ^2 x+2 \cos ^2 y-2}{2 \cos x \cdot \cos y}$
$\cos ^2 x \cdot \cos y=\cos ^2 x+\cos ^2 y-1$
$\cos ^2 x(\cos y-1)=\left(\cos ^2 y-1\right)$
$\cos ^2 x(1-\cos y)=\left(1-\cos ^2 y\right)$
$\cos ^2 x\left(2 \sin ^2 \frac{y}{2}\right)=\sin ^2 y$
$\cos ^2 x\left(2 \sin ^2 \frac{y}{2}\right)=\left(2 \sin \frac{y}{2} \cdot \cos \frac{y}{2}\right)^2$
$\cos ^2 x\left(2 \sin ^2 \frac{y}{2}\right)=4 \sin ^2 \frac{y}{2} \cdot \cos ^2 \frac{y}{2}$
$\cos ^2 x=\left(2 \cos ^2 \frac{y}{2}-1\right)+1$
$\cos y+1=\cos ^2 x$
or $\quad 1+\cos y=\cos ^2 x$
Then, $\cos x=\frac{2 \cos (x-y) \cos (x+y)}{\cos (x+y)+\cos (x-y)}$
$\cos x=\frac{\cos 2 x+\cos 2 y}{2 \cos x \cdot \cos y}$
$\cos x=\frac{2 \cos ^2 x+2 \cos ^2 y-2}{2 \cos x \cdot \cos y}$
$\cos ^2 x \cdot \cos y=\cos ^2 x+\cos ^2 y-1$
$\cos ^2 x(\cos y-1)=\left(\cos ^2 y-1\right)$
$\cos ^2 x(1-\cos y)=\left(1-\cos ^2 y\right)$
$\cos ^2 x\left(2 \sin ^2 \frac{y}{2}\right)=\sin ^2 y$
$\cos ^2 x\left(2 \sin ^2 \frac{y}{2}\right)=\left(2 \sin \frac{y}{2} \cdot \cos \frac{y}{2}\right)^2$
$\cos ^2 x\left(2 \sin ^2 \frac{y}{2}\right)=4 \sin ^2 \frac{y}{2} \cdot \cos ^2 \frac{y}{2}$
$\cos ^2 x=\left(2 \cos ^2 \frac{y}{2}-1\right)+1$
$\cos y+1=\cos ^2 x$
or $\quad 1+\cos y=\cos ^2 x$
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