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This problem is based on this Facebook post.

One can find the value of $x$ in this diagram

enter image description here

by noticing that $\angle CBD=50^{\large\circ}$, and therefore, $$ \frac{\tan\left(10^{\large\circ}\right)}{\tan\left(20^{\large\circ}\right)} =\frac{ED}{CD}=\frac{\tan(x)}{\tan\left(50^{\large\circ}\right)}\tag1 $$ Solving equation $(1)$ gives $$ \tan(x)=\frac{\tan\left(10^{\large\circ}\right)\tan\left(50^{\large\circ}\right)}{\tan\left(20^{\large\circ}\right)}\tag2 $$ Numerically computing the arctangent of the quantity in $(2)$ gives $x=30^{\large\circ}$. This surprised me; I had expected some odd angle, but apparently, this turns out to be a nice angle.

My question is: why does $\tan\left(30^{\large\circ}\right)=\frac{\tan\left(10^{\large\circ}\right)\tan\left(50^{\large\circ}\right)}{\tan\left(20^{\large\circ}\right)}$ ?

robjohn
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  • When this question got a downvote, I looked and saw that the Facebook post had been removed from public view (or was deleted). However, I don't think the Facebook post is an integral part of the question. This is a self-answered question, so I believe it has sufficient context regardless. – robjohn Feb 21 '21 at 09:27
  • Oddly, a very similar question has recently been asked here with an answer of mine there where I use a method completely similar to yours, ending in a similar expression involving tangents of multiples of $10°$ ; by chance, I found a connection to an avatar of the so-called Dorrie's rule (a name coined by Feynman !) with a nice geometric proof in the corresponding Wikipedia article. – Jean Marie Feb 05 '23 at 23:01

3 Answers3

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Answer to the Question $$ \begin{align} \frac{\tan\left(10^{\large\circ}\right)\tan\left(50^{\large\circ}\right)}{\tan\left(20^{\large\circ}\right)} &=\frac1{\tan\left(20^{\large\circ}\right)} \overbrace{\frac{\tan\left(30^{\large\circ}\right)-\tan\left(20^{\large\circ}\right)}{1+\tan\left(30^{\large\circ}\right)\tan\left(20^{\large\circ}\right)}}^{\tan\left(10^{\large\circ}\right)} \overbrace{\frac{\tan\left(30^{\large\circ}\right)+\tan\left(20^{\large\circ}\right)}{1-\tan\left(30^{\large\circ}\right)\tan\left(20^{\large\circ}\right)}}^{\tan\left(50^{\large\circ}\right)}\tag1\\ &=\frac{\frac13-\tan^2\left(20^{\large\circ}\right)}{\tan\left(20^{\large\circ}\right)-\frac13\tan^3\left(20^{\large\circ}\right)}\tag2\\ &=\frac1{\tan\left(60^{\large\circ}\right)}\tag3\\[6pt] &=\tan\left(30^{\large\circ}\right)\tag4 \end{align} $$ Explanation:
$(1)$: $\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$
$(2)$: $\tan\left(30^{\large\circ}\right)=\frac1{\sqrt3}$
$(3)$: $\tan(3x)=\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}$
$(4)$: $\frac1{\tan\left(60^{\large\circ}\right)}=\tan\left(90^{\large\circ}-60^{\large\circ}\right)$

General Identity

The answer above is a special case of the identity $$ \cot(3x)=\frac{\tan\left(\frac\pi6-x\right)\tan\left(\frac\pi6+x\right)}{\tan(x)}\tag5 $$ whose proof mirrors the answer above.

robjohn
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  • $(\tan x)(\tan (\frac{\pi}{3} - x))(\tan \frac{\pi}{3} - x) = \tan 3x$. $x=\frac{\pi}{18}= 10^{\circ} $ in this case –  Jan 10 '21 at 16:54
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    @Forester: $\tan(x)\tan\left(\frac\pi3-x\right)\tan\left(\frac\pi3\color{#C00}{+}x\right)=\tan(3x)$ is the reciprocal of $(5)$ after using $\cot\left(\frac\pi2-x\right)=\tan(x)$ – robjohn Jan 10 '21 at 20:05
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\begin{align} \frac{\tan10\tan50}{\tan20} &= \frac{\sin10\sin50\cos20}{\cos10\cos50\sin20} = \frac{\sin50\cos20}{2\cos^210\cos50}\\ &= \frac{2\cos40\cos20}{2\cos50(1+\cos20)} = \frac{\cos20+\frac12}{2\cos50+\cos70+\frac{\sqrt3}2}\\ &= \frac{\cos20+\frac12}{\cos50+\cos10+\frac{\sqrt3}2} = \frac{\cos20+\frac12}{\sqrt3\cos20+\frac{\sqrt3}2}\\ &= \frac1{\sqrt3}=\tan30 \end{align}

Quanto
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The task is to show that $\tan20^\circ\tan30^\circ=\tan10^\circ\tan50^\circ$. After multiplication through by the cosines of the angles, the task is converted to one of proving that the following quantity is zero:$$a:=\sin20^\circ\sin30^\circ\cos10^\circ\cos50^\circ-\cos20^\circ\cos30^\circ\sin10^\circ\sin50^\circ.$$ Now, from the standard trigonometric formulas $2\cos A\cos B=\cos(A-B)+\cos(A+B)$ and $2\sin A\sin B=\cos(A-B)-\cos(A+B)$, we get $$2a=(\cos10^\circ-\cos50^\circ)\cos10^\circ\cos50^\circ-(\cos10^\circ+\cos50^\circ)\sin10^\circ\sin50^\circ\qquad\qquad\qquad\quad$$ $$=(\cos10^\circ\cos50^\circ-\sin10^\circ\sin50^\circ)\cos10^\circ-(\cos10^\circ\cos50^\circ+\sin10^\circ\sin50^\circ)\cos50^\circ$$ $$=\cos60^\circ\cos10^\circ-\cos40^\circ\cos50^\circ$$ $$\qquad\qquad=\tfrac12\cos10^\circ-\tfrac12(\cos10^\circ+\cos90^\circ)$$ $$=0.$$

John Bentin
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