103 Trigonometry Problems

4. Solutions to Introductory Problems 95
Solution: Note that because of the condition A, B, C ̸= 90 ◦ , all the above
expressions are well defined.
The proof of the identity in part (a) is similar to that of Problem 19(a). By the
arithmetic–geometric means inequality,
By (a), we have
from which (b) follows.
tan A + tan B + tan C ≥ 3 3√ tan A tan B tan C.
tan A tan B tan C ≥ 3 3√ tan A tan B tan C,
Note: Indeed, the identity in (a) holds for all angles A, B, C with A+B +C =
mπ and A, B, C ̸= kπ 2
, where k and m are in Z.
21. Let ABC be a triangle. Prove that
cot A cot B + cot B cot C + cot C cot A = 1.
Conversely, prove that if x,y,z are real numbers with xy + yz+ zx = 1, then
there exists a triangle ABC such that cot A = x, cot B = y, and cot C = z.
Solution: If ABC is a right triangle, then without loss of generality, assume
that A = 90 ◦ . Then cot A = 0 and B + C = 90 ◦ , and so cot B cot C = 1,
implying the desired result.
If A, B, C ̸= 90 ◦ , then tan A tan B tan C is well defined. Multiplying both
sides of the desired identity by tan A tan B tan C reduces the desired result to
Introductory Problem 20(a).
The second claim is true because cot x is a bijective function from the interval
(0 ◦ , 180 ◦ ) to (−∞, ∞).
22. Let ABC be a triangle. Prove that
sin 2 A 2 + B sin2 2 + C sin2 2 + 2 sin A 2 sin B 2 sin C 2 = 1.
Conversely, prove that if x,y,z are positive real numbers such that
x 2 + y 2 + z 2 + 2xyz = 1,