If $ a + b + c = 90^{\circ}$, prove $ \tan(a) \cdot \tan(b) + \tan(b) \cdot \tan(c) + \tan(c) \cdot \tan(a) = 1 $

Question

If $ a + b + c = 90^{\circ}$, prove $ \tan(a) \cdot \tan(b) + \tan(b) \cdot \tan(c) + \tan(c) \cdot \tan(a) = 1 $

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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:

$$ \sin(a+b+c) = 1 \implies \sin(a+b) \cos(c) + \sin(c) \cos(a+b) = 1 $$ $$ \sin(a)\cos(b)\cos(c) + \sin(b)\cos(a)\cos(c) + \sin(c) \cos(a) \cos(b) - \sin(c) \sin(a) \sin(b) = 1$$ $$ \tan(a) \cos(a) \cos(b)\cos(c) + \tan(b) \cos(a) \cos(b) \cos(c) + \tan(c) \cos(a) \cos(b) \cos(c) - \sin(a) \sin(b) \sin(c) = 1$$ $$ \cos(a)\cos(b)\cos(c) (\tan(a) + \tan(b) + \tan(c) - \tan(a)\tan(b)\tan(c)) = 1 $$

Answer 1

Writing $$\tan c=\tan (\pi/2 -(a+b))=\frac 1{tan(a+b)}=\frac {1-\tan a\tan b}{\tan a +\tan b}$$

So we have $$\tan a\tan b+\tan a\tan c+\tan b\tan c=\tan a\tan b+ \frac {\tan a-\tan^2 a\tan b}{\tan a +\tan b}+ \frac {\tan b-\tan a\tan^2 b}{\tan a +\tan b}= \frac {\tan^2 a\tan b+\tan a\tan^2 b+\tan a-\tan^2 a\tan b+\tan b-\tan a\tan^2 b}{\tan a +\tan b}= \frac {\tan a+\tan b}{\tan a +\tan b} =1$$

Answer 2

It's wrong for $a=90^{\circ}$ for example.

But on the domain of our tangents we obtain: $$1-\sum_{cyc}\tan{a}\tan{b}=1-\tan{a}\tan{b}-\tan{c}(\tan{a}+\tan{b})=$$ $$=\left(1-\tan{a}\tan{b}\right)\left(1-\frac{\tan{c}(\tan{a}+\tan{b})}{1-\tan{a}\tan{b}}\right)=$$ $$=\left(1-\tan{a}\tan{b}\right)\left(1-\tan{c}\tan(a+b)\right)=$$ $$=\left(1-\tan{a}\tan{b}\right)\left(1-\tan{c}\cot{c}\right)=0.$$

Answer 3

Since $\tan(a+b+c)$ is undefined, anything where you put it in an equation is suspect. But if you start with $\cot(a+b+c)$ then you can get $1-\tan(a)\tan(b) - \tan(a) \tan(c) -\tan(b) \tan(c)$ in the numerator and show that it must be zero.

I don't see exactly how you intended to finish the proof that starts with $\sin(a+b+c).$ I would start with $\cos(a+b+c)=0$ instead. Expand this using the rule for cosine of a sum twice, then divide by $\cos a \cos b \cos c$ (which must all be non-zero, otherwise the tangents in the formula are not all defined--this really should be an explicit condition of the proposition).

Answer 4

Another concise option: since $\tan c=\frac{1}{\tan(a+b)}=\frac{1-\tan a\tan b}{\tan a+\tan b}$, $\tan b\tan c+\tan c\tan a=1-\tan a\tan b$.