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I'm trying to prove the following problem:

$$\dfrac{\sec^2(x)}{\cot(x)} - \tan^3(x) = \tan(x)$$

I thought I had the answer, but the internet says otherwise. Unfortunately I have been unable to find step-by-step problems to show where I went wrong.

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4 Answers 4

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We will work on the left side and show it is equl to $\tan(x)$. Let's change everything into $\sin(x)$ and $\cos(x)$ using the following formulas $$\sec^2(x) = \frac{1}{\cos^2(x)}, \quad \quad \cot(x) = \frac{\cos(x)}{\sin(x)}, \quad \quad \tan^3(x) = \frac{\sin^3(x)}{\cos^3(x)}.$$ This will give us: $$\frac{\frac{1}{\cos^2(x)}}{\frac{\cos(x)}{\sin(x)}} - \frac{\sin^3(x)}{\cos^3(x)}.$$ Cleaning up the left fraction will give you $$\frac{\sin(x)}{\cos^3(x)} - \frac{\sin^3(x)}{\cos^3(x)}.$$ From here, factor out $\sin(x)$ from the top to get: $$\frac{\sin(x)(1 - \sin^2(x))}{\cos^3(x)}.$$ Finally, use the Pythagorean Identity $\cos^2(x) = 1 - \sin^2(x)$ to finish it off.

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  • $\begingroup$ Thanks for the explanation, but tan(x) being the answer wasn't given; that's what I found online when I went to go see if I was right. So I am just trying to solve it using just the lest side. $\endgroup$
    – Tanner
    Mar 3, 2014 at 3:30
  • $\begingroup$ Oh ok. I will change my answer to reflect that method. $\endgroup$
    – josh
    Mar 3, 2014 at 3:34
  • $\begingroup$ Thanks for changing it, I figured out where I went wrong! (I tried to cancel the cosines in the second step instead of multiplying) $\endgroup$
    – Tanner
    Mar 3, 2014 at 3:58
  • $\begingroup$ @Tanner You are welcome! I'm glad you figured out what went wrong. $\endgroup$
    – josh
    Mar 3, 2014 at 3:59
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Put it all in Tan functions.

$$\sec^2x+1 = tan^2x$$

$$\frac{1}{cotx} =tanx$$

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Hint: $\sec^2 x + 1 = \tan^2 x$. Start with the left side and simplify using this identity.

If you'd like feedback regarding where you went wrong (or for us to check your work), feel free to edit your post to include your solution!

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Hint: $\displaystyle \frac{1}{\cot x} = \tan x$. So we get $\tan x \sec^2 x - \tan^3 x = \tan x$. Divide the equation by a certain trigonometric function and you should get a familiar equation.

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  • $\begingroup$ When I try this I get to $tan(x)(sec(x) + tan(x))(sec(x) - tan(x))$, then I get stuck. Is there a trick to getting around this, or did I go the wrong way? $\endgroup$
    – Tanner
    Mar 3, 2014 at 3:32
  • $\begingroup$ Try using the fact that $\sec^2 x - \tan^2 x = 1$. $\endgroup$
    – 2012ssohn
    Mar 3, 2014 at 5:24

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