What's the shortest path to simplify $\tan{85°} \tan{35°} \tan{15°}+\tan{55°} \tan{5°} \tan{75°}+2\cot{50°}$? We have this expression:
$$\tan{85°} \cdot \tan{35°} \cdot \tan{15°}+\tan{55°} \cdot \tan{5°} \cdot \tan{75°}+2\cot{50°}$$
I want this expression to be in its simplest form.
I fed this into the Symbolab calculator but it gave a very long answer. According to my own information, I wrote it like this:
$$\cot{5°} \cdot \cot{55°} \cdot \cot{75°}+\tan{55°} \cdot \tan{5°} \cdot \tan{75°} + 2\cot{50°}=\\ \\ \frac{\cos{5°} \cdot \cos{55°} \cdot \cos{75°}}{\sin{5°} \cdot \sin{55°} \cdot \sin{75°}}+\frac{\sin{5°} \cdot \sin{55°} \cdot \sin{75°}}{\cos{5°} \cdot \cos{55°} \cdot \cos{75°}}+2\frac{\cos{50°}}{\sin{50°}}$$
And now, if: $$a=\cos{5°} \cdot \cos{55°} \cdot \cos{75°}\\b=\sin{5°} \cdot \sin{55°} \cdot \sin{75°}\\c=\cos{50°}\\d=\sin{50°}$$
It will be very long so i show my final way with the variables $a,b,c,d$:
$$\frac{a}{b}+\frac{b}{a}+2\frac{c}{d}=\frac{a^{2}d+b^{2}d+2abd}{abd}$$
but as you see, it is also very long. I am looking for the simplest form for this expression.

Consider the identity $\tan(x)\cdot\tan(60°-x)\cdot\tan(60°+x)=\tan(3x)$. We now have
$$\tan5°\cdot\tan55°\cdot\tan65°=\tan15°\implies\tan5°\cdot\tan55°\cdot\tan75°=\cot65°$$
and
$$\tan25°\cdot\tan35°\cdot\tan85°=\tan75°\implies\tan15°\cdot\tan35°\cdot\tan85°=\cot25°$$
The expression now is simplified into $$\tan25°+\frac1{\tan25°}+\frac2{\tan50°}=\frac{\tan^225°+1}{\tan25°}+\frac{2}{\tan50°}$$
Using the fact that $\sin(2x)=\frac{2\tan(x)}{1+\tan^2(x)}$, we simplify as
$$\frac{\tan^225°+1}{\tan25°}+\frac{2}{\tan50°}=\frac2{\sin50°}+\frac{2\cos50°}{\sin50°}$$
From the double angle formulae,
$$\frac{2(1+\cos50°)}{\sin50°}=2\cdot\frac{2\cos^225°}{2\sin25°\cos25°}=2\tan65°$$
Therefore,$$\boxed{\tan{85°} \cdot \tan{35°} \cdot \tan{15°}+\tan{55°} \cdot \tan{5°} \cdot \tan{75°}+2\cot{50°}=2\tan65°}$$
