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If you input the trig identity: $$\cot (x)+\tan(x)=\csc(x)\sec(x)$$ Into WolframAlpha, it gives the following proof:

Expand into basic trigonometric parts: $$\frac{\cos(x)}{\sin(x)} + \frac{\sin(x)}{\cos(x)} \stackrel{?}{=} \frac{1}{\sin(x)\cos(x)}$$ Put over a common denominator:

$$\frac{\cos^2(x)+\sin^2(x)}{\cos(x)\sin(x)} \stackrel{?}{=} \frac{1}{\sin(x)\cos(x)}$$

Use the Pythagorean identity $\cos^2(x)+\sin^2(x)=1$:

$$\frac{1}{\sin(x)\cos(x)} \stackrel{?}{=} \frac{1}{\sin(x)\cos(x)}$$

And finally simplify into

$$1\stackrel{?}{=} 1$$

The left and right side are identical, so the identity has been verified.

However, I take some issue with this. All this is doing is manipulating a statement that we don't know the veracity of into a true statement. And I've learned that any false statement can prove any true statement, so if this identity was wrong you could also reduce it to a true statement.

Obviously, this proof can be easily adapted into a proof by simply manipulating one side into the other, but:

Is this proof correct on its own? And can the steps WolframAlpha takes be justified, or is it completely wrong?

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    $\begingroup$ Can the downvoter explain their reasoning? $\endgroup$
    – Nico A
    Dec 7, 2017 at 15:44
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    $\begingroup$ You are correct that this sort of proof is wrong - you really need to go the other direction. In this case, you can (assuming $\cos x\neq 0$ and $\sin x\neq 0$. If either is zero, then both sides of the original equation are undefined.) $\endgroup$
    – Thomas Andrews
    Dec 7, 2017 at 15:44
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    $\begingroup$ You can read the proof as saying that each identity holds if and only if the following one holds. Since the last identity is indeed true, the result follows. Whether additional short comments need to be added to verify that indeed each new potential identity is equivalent to the preceding one probably depends on the precise phrasing used by Alpha. $\endgroup$
    – Andrés E. Caicedo
    Dec 7, 2017 at 15:44
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    $\begingroup$ Well, I suppose how the proof is phrased, I guess, but unless the writer says "if and only if." This kind of proof is risky because it moves in one direction. (It is not true that if $1=1$ then the previous equation is true, for example, unless $\cos x\sin x\neq 0$. You'd spot that error if you wrote the proof in the reverse.) @AndresMejia And there are many cases of "fake proofs" that look like this where reverse steps don't work, so it is a valid complaint for this proof, as well. $\endgroup$
    – Thomas Andrews
    Dec 7, 2017 at 15:51
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    $\begingroup$ @RolazaroAzeveires By showing a statement to be equivalent to a true statement, or manipulating a true statement into our statement of unknown veracity. But starting with a statement that we don't know the truth value of, it could be false, and a false statement implies any true statement. $\endgroup$
    – Nico A
    Dec 7, 2017 at 22:50

2 Answers 2

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It is good that you are wary of proving identities this way. Indeed, I could "prove" $0=1$ by saying

\begin{align*} 0 &\stackrel{?}{=}1\\ 0\cdot 0 &\stackrel{?}{=} 0 \cdot 1\\ 0 &=0. \end{align*}

The important point is that every step WolframAlpha did is reversible, while the step I took (multiplying by $0$) was not. That is what allows the proof from WolframAlpha to be rearranged into a proof that starts with one side of the identity and ends at the other:

\begin{align*} \cot(x)+\tan(x) &= \frac{\cos(x)}{\sin(x)} + \frac{\sin(x)}{\cos(x)}\\ &= \frac{\cos^2(x)}{\sin(x)\cos(x)} + \frac{\sin^2(x)}{\sin(x)\cos(x)}\\ &= \frac{\sin^2(x)+\cos^2(x)}{\sin(x)\cos(x)}\\ &=\frac{1}{\sin(x)\cos(x)}\\ &=\csc(x)\sec(x). \end{align*}

So no, the WolframAlpha proof is not wrong, but it neglects to emphasize the important fact that every step is reversible. I am not a fan of that sort of proof, as it gives students the idea that they can prove an identity by manipulating both sides in any way they like to arrive at a true statement.

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    $\begingroup$ +1, I think that they should edit the page just for pedagogical reasons. $\endgroup$
    – Andres Mejia
    Dec 7, 2017 at 15:50
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    $\begingroup$ +1: it's absurd and counter-intuitive to write the chain of equalities the way WolframAlpha does. $\endgroup$
    – Rob Arthan
    Dec 7, 2017 at 19:16
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    $\begingroup$ @RobArthan I would argue that it's absurd, but not counter-intuitive. If trying to prove such an identity, it's perfectly reasonable to begin from the statement you wish to prove and, taking care that all steps are reversible, try to manipulate it into something you know is true. $\endgroup$
    – Carl Schildkraut
    Dec 7, 2017 at 21:29
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    $\begingroup$ I would say that it is neither absurd nor counter-intuitive. That's exactly how you find proofs for such things. $\endgroup$
    – tomasz
    Dec 7, 2017 at 21:39
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    $\begingroup$ @AndresMejia: I don't think this page is a human generated wiki. It's a dynamic step-by-step "proof" done automatically by WolframAlpha. $\endgroup$
    – Eric Duminil
    Dec 8, 2017 at 8:38
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You can make it rigorous by going in the reverse direction. $1=1 \implies \frac{1}{\cos x \sin x}=\frac{1}{\cos x \sin x} \implies \dots$.

But this is a silly looking "proof" and it is really clunky. The point is that this is not $X \implies 1=1$, but rather the proof given ensures that $X \iff 1=1$ by following equality (which is symmetric) in either direction for the proof.

Personally the way you suggest is much preferred , $$\frac{1}{\cos x \sin x}=\frac{\cos^2x+\sin^2x}{\cos x\sin x}=\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}=\cot x+\tan x.$$

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    $\begingroup$ "ommitted x everywhere" why? Do we need such new notation? $\endgroup$
    – Rolazaro Azeveires
    Dec 7, 2017 at 21:47
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    $\begingroup$ Of course not, I was just in a rush, and I don’t think the argument suffers from a slight abbreviation. $\endgroup$
    – Andres Mejia
    Dec 7, 2017 at 22:01
  • $\begingroup$ Sure, the answer was still clear, and that is something I would scribble with pencil and papper. But explainig you "ommitted x everywhere", 19 characters. Ommited x's, 16. So it felt like an option on the notation. $\endgroup$
    – Rolazaro Azeveires
    Dec 8, 2017 at 11:26

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