MHB Can You Solve the Trigonometry Puzzle of the Week?

[anemone]

Here is this week's POTW:

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Prove that $\cot 13^\circ \cot 23^\circ \tan 31^\circ \tan 35^\circ \tan 41^\circ =\tan 75^\circ$

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to kaliprasad for his correct solution!:)

Here's the suggested model solution:

Spoiler

First notice that the given equality can be rewritten as $(\tan 31^\circ \tan 41^\circ \tan 67^\circ \tan 77^\circ)\tan 35^\circ=\tan 75^\circ$(**)

Next, observe that $$\tan(5\cdot31^\circ)=\tan155^\circ=\tan(180^\circ-25^\circ)=-\tan25^\circ$$.

This works for $41^\circ,\,67^\circ$ and $77^\circ$.

Moreover, $$\tan(5\cdot(-5)^\circ)=-\tan25^\circ$$, so we get $\tan 5x=-\tan 25^\circ$ for $x\in \{-5,\,31,\,41,\,67,\,77\}$. Therefore, putting these $x$s into the identity $$\tan 5x=\frac{5\tan x-10\tan^3 x+\tan^5 x}{1-10\tan^2 x+5\tan^4 x}$$ and rearranging we get that $\tan x^\circ$ is a root of the polynomial $t^5+5\tan25^\circ t^4-10t^3-10\tan 25^\circ t^2+5t+\tan 25^\circ$ for $x\in \{-5,\,31,\,41,\,67,\,77\}$.

Hence by Viete's relations, we have

$\tan (-5^\circ)^\circ \tan31^\circ \tan 41^\circ \tan 67^\circ \tan 77^\circ =-\tan 25^\circ$, or equivalently,

$ \tan 31^\circ \tan 41^\circ \tan 67^\circ \tan 77^\circ=\tan 25^\circ\tan 85^\circ$

But we know that $\tan 35^\circ \tan 25^\circ \tan 85^\circ=\tan 75^\circ$ therefore we have proved that

$(\tan 31^\circ \tan 41^\circ \tan 67^\circ \tan 77^\circ)\tan 35^\circ=\tan 75^\circ$, i.e.

$\cot 13^\circ \cot 23^\circ \tan 31^\circ \tan 35^\circ \tan 41^\circ =\tan 75^\circ$