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1 solution
Let C = π − ( A + B ) so that t a n ( C ) = t a n ( π − ( A + B ) ) = 1 + t a n ( π ) t a n ( A + B ) t a n ( π ) − t a n ( A + B ) = − t a n ( A + B ) . Now turning our attention to the given numerator:
N = t a n ( A ) + t a n ( B ) + t a n ( C ) ⇒ t a n ( A ) + t a n ( B ) − t a n ( A + B ) = t a n ( A ) + t a n ( B ) − 1 − t a n ( A ) t a n ( B ) t a n ( A ) + t a n ( B ) ;
or 1 − t a n ( A ) t a n ( B ) [ t a n ( A ) + t a n ( B ) ] [ 1 − t a n ( A ) t a n ( B ) ] + [ t a n ( A ) + t a n ( B ) ] ;
or 1 − t a n ( A ) t a n ( B ) t a n ( A ) + t a n ( B ) − t a n ( A ) − t a n ( B ) − t a n 2 ( A ) t a n ( B ) − t a n ( A ) t a n 2 ( B ) ;
or N = − 1 − t a n ( A ) t a n ( B ) t a n 2 ( A ) t a n ( B ) + t a n ( A ) t a n 2 ( B ) .
Now for the denominator:
D = t a n ( A ) t a n ( B ) t a n ( C ) = t a n ( A ) t a n ( B ) [ − t a n ( A + B ) ] ;
or − t a n ( A ) t a n ( B ) ⋅ 1 − t a n ( A ) t a n ( B ) t a n ( A ) + t a n ( B ) ;
or D = − 1 − t a n ( A ) t a n ( B ) t a n 2 ( A ) t a n ( B ) + t a n ( A ) t a n 2 ( B ) .
Since N = D , the original ratio equals unity ⇒ t a n ( A ) t a n ( B ) t a n ( C ) t a n ( A ) + t a n ( B ) + t a n ( C ) = 1 for any triangle △ A B C .