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$$ \left(\tan^2x +\frac{1}{\tan x}\right) \left(\sin x+\cos x\right)=\left(\frac{1}{\cos x}\right)+\left(\frac{1}{sin x}\right) $$

Can someone show me how to solve this identity and also explain the steps?

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    $\begingroup$ You could try expressing everything in terms of sines and cosines, clear fractions, and see what you get. $\endgroup$ – Gerry Myerson Dec 4 '13 at 2:14
  • $\begingroup$ I don't think this is true. See Wolfram $\endgroup$ – Calvin Lin Dec 4 '13 at 2:16
  • $\begingroup$ It's false. Plug in $x=\pi/4$. $\endgroup$ – Doc Dec 4 '13 at 2:23
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    $\begingroup$ Well, did you try to follow Gerry's suggestion above? $\endgroup$ – Doc Dec 4 '13 at 2:46
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    $\begingroup$ Bet you anything you want that your $\tan^2{x}$ term should be $\tan{x}$. $\endgroup$ – Doc Dec 4 '13 at 2:51
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Multiplying out on the left hand side you get: $(tan + \frac{1}{tan})\cdot (sin+cos) \\ = (\frac{sin}{cos}+\frac{cos}{sin})\cdot (sin + cos) \\ = \frac{sin^2}{cos}+ cos + sin + \frac{cos^2}{sin} \\ = \frac{sin^2}{cos}+\frac{cos^2}{cos}+\frac{sin^2}{sin}+\frac{cos^2}{sin} \\ = \frac{1}{cos}+\frac{1}{sin}.$

Hence, $(tan + \frac{1}{tan})\cdot (sin+cos) = \frac{1}{cos}+\frac{1}{sin}$. Now just apply these operators to $x$.

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    $\begingroup$ Depriving OP of the joy of working this out on his/her own. $\endgroup$ – Gerry Myerson Dec 4 '13 at 4:44
  • $\begingroup$ @GerryMyerson, perhaps... I was feeling generous today. $\endgroup$ – Chris K Dec 4 '13 at 4:55
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    $\begingroup$ I don't follow. The generous thing to do is to let OP work through the answer. $\endgroup$ – Gerry Myerson Dec 4 '13 at 5:12
  • $\begingroup$ @GerryMyerson: In retrospect, I probably should not have answered; will hold off from answering questions of a similar ilk in future. $\endgroup$ – Chris K Dec 4 '13 at 22:49

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