原式=tan10°×tan20°×tan30°×tan40°×tan50°×tan60°×tan70°×tan80°
=(tan10°×tan80°)×(tan20°×tan70°)×(tan30°×tan60°)×(tan40°×tan50°)
=(tan10°×cot10°)×(tan20°×cot20°)×(tan30°×cot30°)×(tan40°×cot40°)
=1
(2) m=n
(3) 由題意,可知
cosA+cosB=(m+1)/2,cosA*cosB=m/4
即 2cos(A/2+B/2)*cos(A/2-B/2)=(m+1)/2
(1/2)*[cos(A+B)+cos(A-B)]=m/4
則 sin(C/2)*cos(A/2-B/2)=(m+1)/4
-cosC+cos(A-B)=m/2
又 ∠C=90°
故 cos(A/2-B/2)=√2(m+1)/4
cos(A-B)=m/2
所以 m/2=2[√2(m+1)/4]²-1
整理,得
m²-3=0
解,得
m=±√3