How can I find the following product using elementary trigonometry?

$$ \tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdots \tan 89^\circ.$$

I have tried using a substitution, but nothing has worked.


HINT : $$ \tan x\cdot\tan(90^\circ-x)=1. $$

  • 4
    $\begingroup$ Also $\cot x=\tan(90^\circ-x)$ and $\cot x=\dfrac{1}{\tan x}$. $\endgroup$ – Tunk-Fey May 22 '14 at 18:31

Hint: $\tan a^{\circ} = \cot (90-a)^{\circ}$.

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    $\begingroup$ Why the downvote? $\endgroup$ – rogerl Aug 25 '14 at 21:58

First, let's re-arrange these terms so that we can make use of the hints in other answers.

$$\tan(1^\circ) \cdot \tan(89^\circ) \cdot \tan(2^\circ) \cdot \tan(88^\circ) \cdot\cdot\cdot \tan(44^\circ) \cdot \tan(46^\circ) \cdot \tan(45^\circ)$$

Here, we can see a clear pattern of $$\tan(x) \cdot\tan(90^\circ-x)$$ repeating, except for 45, who has no dance partner.

Now, we can use the fact that $$\tan(x) \cdot \tan(90^\circ - x) = 1$$ and reduce all the pairs of numbers to 1. We're left with $$ 1 \cdot 1 \cdot 1 \cdot \cdot \cdot \tan(45^\circ)$$

and since $$ tan(45^\circ) = 1 $$

we get an answer of $$\tan(1^\circ) \cdot \tan(2^\circ) \cdot \tan(3^\circ) \cdot\cdot\cdot \tan(89^\circ) = 1$$


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