• gold
prove: sinx+cosx/tan^2x-1 = cos^2x/sinx-cosx
  • Stacey Warren - Expert
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  • schrodinger
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  • gold
\[\sin ^{4}x - \cos ^{4}x/ \tan ^{4}x -1 = \cos ^2x\]
  • shadowfiend
I'm going to look at the first equation for now. Is the second a different proof that you need to do? There are two things to keep in mind here. First off, remember when trying to prove these that you can work both sides (it will be easier). Secondly, remember that \(\tan x = \frac{\sin x}{\cos x}\), which means that \(\tan^2 x = \frac{\sin^2 x}{\cos^2 x}\). So, once we've determined that, we can first take \(\sin x - \cos x\) from the right and multiply it into the left: \[\begin{align} \frac{\sin x + \cos x}{\tan^2 x - 1} &= \frac{\cos^2 x}{\sin x - \cos x}\\ \frac{(\sin x + \cos x)(\sin x - \cos x)}{\tan^2 x - 1} &= \cos^2 x\\ \frac{\sin^2 x - \cos^2 x}{\tan^2 x - 1} &= \cos^2 x \end{align}\] You can then take \(\tan^2 x - 1\) and multiply it into the right: \[\begin{align} \frac{\sin^2 x - \cos^2 x}{\tan^2 x - 1} &= \cos^2 x\\ \sin^2 x - \cos^2 x &= \cos^2 x(\tan^2 x - 1) \end{align}\] Then we use the fact that \(\tan^2 x = \frac{\sin^2 x}{\cos^2 x}\) and multiply through by \(\cos^2 x\) on the right: \[\begin{align} \sin^2 x - \cos^2 x &= \cos^2 x(\tan^2 x - 1)\\ &=\cos^2 x \tan^2 x - \cos^2 x\\ &= \cos^2 x\frac{\sin^2 x}{\cos^2 x} - \cos^2 x\\ \sin^2 x - \cos^2 x = \sin^2 x - \cos^2 x \end{align}\] When we expand \(\tan^2 x\), we cancel out the \(\cos^2 x\) and we are left with the same expressions on both sides, which means we have proven the identity.

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