<s>[INST] If $ a + b + c = 90^{\circ}$, prove $ \tan(a) \cdot \tan(b) + \tan(b) \cdot \tan(c) + \tan(c) \cdot \tan(a) = 1 $ If $ a + b + c = 90^{\circ}$, prove  $ \tan(a) \cdot \tan(b) + \tan(b) \cdot \tan(c) + \tan(c) \cdot \tan(a) = 1 $


Attempt:
Notice that $$ \tan(a+b+c) = \frac{\tan(a+b) + \tan(c)}{1 - \tan(a+b)\tan(c)} $$
$$ = \frac{\frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)} + \tan(c)}{1 - \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)} \tan(c)} $$
$$ = \frac{\tan(a) + \tan(b) + \tan(c) - \tan(a) \tan(b) \tan(c)}{ (1-\tan(a)\tan(b)) - \tan(a) \tan(c) -\tan(b) \tan(c) } $$
Then the denomenator must be $0$, so it is proven?
Another way:
$$ \sin(a+b+c) = 1 \implies \sin(a+b) \cos(c) + \sin(c) \cos(a+b) = 1  $$
$$ \sin(a)\cos(b)\cos(c) + \sin(b)\cos(a)\cos(c) + \sin(c) \cos(a) \cos(b) - \sin(c) \sin(a) \sin(b) = 1$$
$$ \tan(a) \cos(a) \cos(b)\cos(c) + \tan(b) \cos(a) \cos(b) \cos(c) + \tan(c) \cos(a) \cos(b) \cos(c) - \sin(a) \sin(b) \sin(c) = 1$$
$$ \cos(a)\cos(b)\cos(c) (\tan(a) + \tan(b) + \tan(c) - \tan(a)\tan(b)\tan(c)) = 1 $$
 [/INST] It's wrong for $a=90^{\circ}$ for example.
But on the domain of our tangents we obtain:
$$1-\sum_{cyc}\tan{a}\tan{b}=1-\tan{a}\tan{b}-\tan{c}(\tan{a}+\tan{b})=$$
$$=\left(1-\tan{a}\tan{b}\right)\left(1-\frac{\tan{c}(\tan{a}+\tan{b})}{1-\tan{a}\tan{b}}\right)=$$
$$=\left(1-\tan{a}\tan{b}\right)\left(1-\tan{c}\tan(a+b)\right)=$$
$$=\left(1-\tan{a}\tan{b}\right)\left(1-\tan{c}\cot{c}\right)=0.$$
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