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Prove that $$\frac{(2\cos^2 x - 1)^2}{\cos^4 x - \sin^4 x} = 1 - 2 \sin^2x$$

Thanks!

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    $\begingroup$ Please show your work and where you faced problems $\endgroup$ – Vizag Apr 20 '19 at 17:16
  • $\begingroup$ Are you able to solve now? Or are you still facing problems? $\endgroup$ – Vizag Apr 20 '19 at 18:34
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Hint: $$2\cos^{2}{(x)} - 1 = \cos{(2x)} = \cos^{2}{(x)} -\sin^{2}{(x)}$$

Can you take it from here?

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In other words, $(\cos 2x)^2/\cos 2x=\cos 2x$, where one of the three famous expressions for $\cos 2x$ has been multiplied by $\cos^2x+\sin^2x=1$ to throw you off the scent. (As @OscarLanzi points out, we could instead think of them as three expressions for $\cos^2x-\sin^2x$ if we didn't know double angle formulae.)

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  • $\begingroup$ You don't need the double angle formulas at all. They throw the scent off in this problem. $\endgroup$ – Oscar Lanzi Apr 20 '19 at 17:22
  • $\begingroup$ @OscarLanzi Thanks; edited. $\endgroup$ – J.G. Apr 20 '19 at 17:32
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The denominator is $(\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x)$. Furthermore, you should know something about $\cos^2 x + \sin^2 x$...

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  1. Factor $\cos^4 x-\sin^4 x=(\cos^2 x)^2-(\sin^2 x)^2$ as a difference of squares.

  2. Render $\cos^2 x+\sin^2 x = 1$ to eliminate the cosine terms.

  3. Everything falls into place.

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