已知sin2(a+r)=nsin2b,则tan(a+b+c)/tan(a-b+r)=An-1/n+1Bn/n+1Cn/n-1Dn+1/n-1求详解.

Question

已知sin2(a+r)=nsin2b,则tan(a+b+c)/tan(a-b+r)=
A n-1/n+1
B n/n+1
C n/n-1
D n+1/n-1
求详解.

Answer 1

sin2(a+r)=nsin2b
∵sin[2(a+r)]=sin[(a+b+r)+(a-b+r)]=sin(a+b+r)cos(a-b+r)+cos(a+b+r)sin(a-b+r)
sin2b=sin[(a+b+r)-(a-b+r)]=sin(a+b+r)cos(a-b+r)-cos(a+b+r)sin(a-b+r)
∴sin(a+b+r)cos(a-b+r)+cos(a+b+r)sin(a-b+r)=n[sin(a+b+r)cos(a-b+r)-cos(a+b+r)sin(a-b+r)]
∴(n-1)sin(a+b+r)cos(a-b+r)=（n+1)cos(a+b+r)sin(a-b+r)
∴sin(a+b+r)/cos(a+b+r)*cos(a-b+r)/sin(a-b+r)=(n+1)/(n-1)
∴tan(a+b+c)/tan(a-b+r)=（n+1)/(n-1)
选D