If ( x - y ) , x and ( x + y ) are in HP , then x ( y / 2 ) is...

Given that, $\cos ( x - y ) , \cos x, \cos ( x + y )$ are in $HP$.

Then, $\cos x = \dfrac{2\cos ( x - y ) \cos ( x + y )}{\cos ( x - y ) + \cos ( x + y )}$

$⇒ \cos x = \dfrac{2\left(\cos^{2}x - \sin^{2}y\right)}{2\cos x \cos y}$

$⇒ \cos^{2}xcosy = \cos^{2}x - \sin^{2}y$

$⇒ \cos^{2}x ( \cos y - 1 ) = - \sin^{2}y$

$⇒ \cos^{2}x ( 1 - \cos y ) = 1 - \cos^{2}y$

$⇒ \cos^{2}x ( 1 - \cos y ) = ( 1 - \cos y ) ( 1 + \cos y )$

$⇒ \cos^{2}x = 1 + \cos y ⇒ \cos^{2}x = 2\cos^{2}y / 2$

$⇒ \cos^{2}x\sec^{2} ( y / 2 ) = 2$

$⇒ \cos x \sec ( y / 2 ) = \pm \sqrt{2}$