Tough trigonometric identity: $\cot 13^\circ\cot 23^\circ \tan 31^\circ\tan35^\circ\tan41^\circ = \tan 75^\circ$

Question

Prove that $$\cot 13^o\cot 23^o \tan 31^o\tan35^o\tan41^o = \tan 75^o$$

I managed to rearrange it to the form $$\tan 31^o\tan 35^o\cot 49^o = \cot 15^o\tan 23^o\cot 77^o$$ and in this form we have the interesting property that the sum of arguments on both sides is equal, i.e. 31+35+49=15+23+77. I couldn't get past this stage, so I would appreciate any help.

EDIT: I found that $\tan x\tan(60^o-x)\tan (60^o+x)=\tan 3x$. Perhaps someone can use this to solve the problem. I myself haven't been able to. Thanks.

Answer 1

First convert all ratios of the left hand side in tangent

Observe that $\tan(5\cdot77^\circ)=\cdots=\tan25^\circ$ etc.

So, let $\tan5x=\tan25^\circ\implies5x=180^\circ n+25^\circ\iff x=36^\circ n+5^\circ$ where $n$ is any integer

Like Sum of tangent functions where arguments are in specific arithmetic series or this

$$\tan5x=\frac{\binom51\tan x-\binom53\tan^3x+\binom55\tan^5x}{\binom50-\binom52\tan^2x+\binom54\tan^4x}$$

If $\tan5x=\tan25^\circ,$

$$\tan^5x-\cdots-\binom50\tan25^\circ=0$$

$$\prod_{r=0}^4\tan\left(36^\circ\cdot r+5^\circ\right)=\frac{\tan25^\circ}1$$

$r=0\implies\tan\left(36^\circ\cdot0+5^\circ\right)=\tan5^\circ$

$r=1\implies\tan\left(36^\circ\cdot1+5^\circ\right)=\tan41^\circ$

$r=2\implies\tan\left(36^\circ\cdot2+5^\circ\right)=\tan77^\circ=\cot13^\circ$ $r=3\implies\tan\left(36^\circ\cdot3+5^\circ\right)=\tan113^\circ=-\tan67^\circ=-\cot23^\circ$ $r=4\implies\tan\left(36^\circ\cdot4+5^\circ\right)=\tan149^\circ=-\tan31^\circ$

So, we need to show $$\tan35^\circ\frac{\tan25^\circ}{\tan5^\circ}=\tan75^\circ$$

which is readily available from your formula putting $x=25^\circ$ mentioned here (How can I find the following product? $ \tan 20^\circ \cdot \tan 40^\circ \cdot \tan 80^\circ.$)

Answer 2

If $$ A = \tan x, \; \; B = \tan y, \; \; C = \tan z, $$ then $$ ABC = A+B+C - (1-BC-CA-AB) \tan(x+y+z). $$ In your case, you need to also be liberal with the use of $$ \cot t = \tan \left( 90^\circ - t \right) $$ to get the same $\tan{x+y+z}$ term for both left and right sides.