103 Trigonometry Problems

94 103 Trigonometry Problems
(b)
tan A 2 tan B 2 tan C 2 ≤ √
3
9 .
Solution: By the addition and subtraction formulas, wehave
tan A 2 + tan B 2 = tan A + B
2
Because A + B + C = 180 ◦ , A+B
2
= 90 ◦ − C 2
Thus,
(
1 − tan A 2 tan B )
.
2
, and so tan
A+B
2
= cot C 2 .
tan A 2 tan B 2 + tan B 2 tan C 2 + tan C 2 tan A 2
= tan A 2 tan B 2 + tan C 2 cot C (
1 − tan A 2 2 tan B )
2
= tan A 2 tan B 2 + 1 − tan A 2 tan B 2 = 1,
establishing (a).
By the arithmetic–geometric means inequality, wehave
1 = tan A 2 tan B 2 + tan B 2 tan C 2 + tan C 2 tan A 2
≥ 3 3 √ (
tan A 2 tan B 2 tan C 2
) 2
,
from which (b) follows.
Note: An equivalent form of (a) is
cot A 2 + cot B 2 + cot C 2 = cot A 2 cot B 2 cot C 2 .
20. Let ABC be an acute-angled triangle. Prove that
(a) tan A + tan B + tan C = tan A tan B tan C;
(b) tan A tan B tan C ≥ 3 √ 3.