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mathslover
 3 years ago
If \(\cos(xy)\) , \(\cos x \) and \(\cos (x+y)\) are in Harmonic progression. then evaluate \(\cos x . \sec (\frac{y}{2})\)
mathslover
 3 years ago
If \(\cos(xy)\) , \(\cos x \) and \(\cos (x+y)\) are in Harmonic progression. then evaluate \(\cos x . \sec (\frac{y}{2})\)

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mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0i tried to do like this first.

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Should I simplify it further?

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Or I am going on wrong track?

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0I have : \(\cfrac{2\cos (x+y) \cos (xy) }{\cos (xy) + \cos (x+y)}= \cos x\) Now : \(2\cos (x+y ) \cos (xy) = \cos x \cos (xy) + \cos x \cos (x+y)\) \(\implies 2\cos^2x \cos^2y  2\sin^2 x \sin^2 y = \cos x ( \cos (x+y) + \cos (xy))\) \(\cfrac{2(\cos^2x \cos^2y  \sin^2x \sin^2y) }{\cos(x+y) + \cos(xy)}= \cos x\) \(\cfrac{\cos^2x \cos^2y  \sin^2x \sin^2y}{\cos x\cos y }= \cos x\) \(\cos^2x\cos^2y  \sin^2x \sin^2y = \cos^2x \cos y\) \(\cos^2x \cos^2y  \cos^2 x \cos y = \sin^2x \sin^2y\) \(\cos^2x \cos^2y  \cos^2 x \cos y = (1\cos^2x)(1\cos^2y)\) \(\cos^2x\cos^2y\cos^2x\cos y = 1  \cos^2y  \cos^2x + \cos^2x\cos^2y\) \(\cancel{cos^2x\cos^2y}  \cos^2x \cos y = 1  \cos^2y  \cos^2x + \cancel{\cos^2x\cos^2y}\) \(\cos^2x\cos y = 1\cos^2y  \cos^2x\) \(1\cos^2y  \cos^2x + \cos^2x \cos y = 0 \) \((1+\cos y)(1+\cos y)  \cos^2x (1  \cos y) = 0\) \((1\cos y)(1+\cos y  \cos^2x ) = 0\) Therefore cos y = 1 and \(\cos x = \sqrt{2}\)

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0So I get :  \(\sqrt{2}\)  = \(\sqrt{2}\) @amistre64 would you please take time to check my method.

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Any help @UnkleRhaukus ?

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0@satellite73 No clue?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it looks like you have done a ton of work, but i cannot do this in a few minutes, would take me at least an hour i can look at it later, it is real math not off the top of my head math

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0No problem but it didn't take me more than 10 minutes. Trust me ... its not that leanthy.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0all though a quick google search tells me your answer is right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and gives a slightly shorter method here is the link but ignore it if you want http://www.qfak.com/education_reference/science_mathematics/?id=b717416#.UUsQ1BesiSo

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0great! Thanks.. A medal deserves
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