Problem
GEO-B3-M06-P015 Tangents of Half-Angles
#15
★★★☆☆ Level 3 of 5
Prove that for the angles of triangle \(ABC\), \(\tan\frac A2\tan\frac B2+\tan\frac B2\tan\frac C2+\tan\frac C2\tan\frac A2=1\).
Inspired by Prasolov trigonometric geometry method
C. Hint 1. Use \(A/2+B/2+C/2=90^\circ\).
D. Hint 2. Apply the tangent formula for a sum of three angles.
Let \(x=\tan\frac A2\), \(y=\tan\frac B2\), \(z=\tan\frac C2\). Since \(\frac A2+\frac B2+\frac C2=90^\circ\), the tangent of the sum is infinite.
Using \(\tan(u+v+w)=\frac{x+y+z-xyz}{1-xy-yz-zx}\), the denominator must be zero. Hence \(1-xy-yz-zx=0\), so \(xy+yz+zx=1\).
The formula is important in problems involving \(r,p,R\).