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If A,B,C are angles of a triangle show that:

$$\tan A+ \tan B+\tan C = \tan A \tan B \tan C $$

I've tried this many times but I cannot seem to prove it, can someone show me how to solve this problem?

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  • $\begingroup$ use the facts that $A+B+C=\pi$, $\tan(\pi-x)=-\tan x$, and the addition formula for tangent $$ \tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y} $$ $\endgroup$ – yoyo May 17 '15 at 1:09
  • $\begingroup$ @FelipeZC: Do you understand? $\endgroup$ – user174622 May 17 '15 at 1:12
  • $\begingroup$ Yeah, thanks man $\endgroup$ – Felipe ZC May 17 '15 at 1:13
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set $t_a=\tan A$ etc.

for any pair of angles $A,B$ we have $$ \tan(A+B) = \frac{t_a+t_b}{1-t_at_b} $$ so for any three angles $A,B,C$: $$ \tan(A+B+C)= \frac{ \frac{t_a+t_b}{1-t_at_b} +t_c}{1-\frac{t_a+t_b}{1-t_at_b}t_c} $$ $$ =\frac{t_a+t_b+t_c -t_at_bt_c}{1-(t_at_b+t_bt_c+t_ct_a)} \tag{1} $$ if, in addition, we know that: $$ A+B+C = n\pi \tag{2} $$ then $$ \tan(A+B+C) = 0 $$ for a planar triangle (2) holds with $n=1$, so in (1) the numerator must be zero, giving: $$ \tan A + \tan B + \tan C = \tan A \tan B \tan C $$

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$$ 0 = \tan 180^\circ = \tan (A+B+C) $$ Apply to the line above the formula for the tangent of a sum of two numbers. You can take the two numbers to be $A$ and $B+C$, so $$ \tan(A+(B+C)) = \frac{\tan A + \tan(B+C)}{1-\tan A\tan(B+C)}. $$ Then apply it again to $\tan(B+C)$. Do the routine simplifications. You get a fraction. A fraction is zero only if the numerator is zero. The identity you're trying to prove just says the numerator of this fraction is zero.

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