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I calculated $$\tan 75^o - [\cot 13^o\cdot \cot 23^o \cdot \tan 31^o \cdot \tan 35^o\cdot \tan41^o]$$ and I got a nonzero answer:
http://www.wolframalpha.com/input/?i=tan%2875%29-[cot%2813%29*cot%2823%29*tan%2831%29*tan%2835%29*tan%2841%29]

But someone posted this problem in a forum as an identity, i.e. $$\tan 75^o = \cot 13^o\cdot \cot 23^o \cdot \tan 31^o \cdot \tan 35^o\cdot \tan 41^o$$

So my question is: Should the answer be $0$? And if so, why is Wolfram Alpha giving a nonzero answer?

Thanks!

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    $\begingroup$ Maybe WolframAlpha interprets the inputs as radians, not degrees. Verify that... $\endgroup$ – cjferes Sep 25 '14 at 19:31
  • $\begingroup$ look at the numberline $\endgroup$ – ganeshie8 Sep 25 '14 at 19:32
  • $\begingroup$ Besides the above explanations, WA is, astonishing (?) enough, not infallible: it makes mistakes, and quite a few, when trying to do some stuff. Be careful with that. $\endgroup$ – Timbuc Sep 25 '14 at 19:33
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    $\begingroup$ @AnalysisIncarnate If Mathematica gives you $0$, then it's (almost certainly) a rounding error, since WA doesn't use the full power of symbolic manipulation and therefore computes the product as a floating point value. $\endgroup$ – Daniel Fischer Sep 25 '14 at 19:39
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    $\begingroup$ Probably not a coincidence. Playing around with the addition theorems may lead to the goal. $\endgroup$ – Daniel Fischer Sep 25 '14 at 19:55
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Try asking it to do it numerically more explicitly (putting in a decimal point makes even mathematica assume you're using approximate numeric values, and so it will use approximate numerical methods) and you'll see that the value you get is well within the default tolerance value of the WA engine. Curiously, you'll note the output it gives for "Result" and "Number line" don't really agree. The result value is of substantially higher precision, and substantially closer to zero in terms of how negative the exponent is (and if you ask it for more digits, you'll get even closer to zero). Presumably it is using different methods for each--lower precision ones for the number line, namely--, but we'd need to know more about the precise workings of these features to know for sure.

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