mathschallenge.net

Problem

Prove that tan$A$ + tan$B$ + tan$C$ = tan$A$ tan$B$ tan$C$ for any non-right angle triangle.

Solution

As $A$ + $B$ + $C$ = 180^o, it follows that tan($A$ + $B$ + $C$) = tan180^o = 0.

Using the addition formula:

tan($A$ + ($B$ + $C$))

=

tan$A$ + tan($B$ + $C$) 1 tan$A$ tan($B$ + $C$)

= 0

tan$A$ + tan($B$ + $C$) = 0.

Using the addition formula again:

tan A +

tan$B$ + tan$C$ 1 tan$B$ tan$C$

= 0

tan$A$(1 tan$B$ tan$C$) + tan$B$ + tan$C$ = 0

tanA tan$A$ tan$B$ tan$C$ + tan$B$ + tan$C$ = 0

Hence, tan$A$ + tan$B$ + tan$C$ = tan$A$ tan$B$ tan$C$

As A + B = 180^o C, use tan(A + B) = tan(180^o C) to prove the same result by a slightly simpler route.