If an angle A is divided into b and c,such that,tanb : tanc=x : y,show that sin(b-c)=x-y/x+y*sinA

Question

help me please

Answer 1

A = B + C,  tanB/tanC = x/y

Starting with (x-y)/(x+y)*sinA,

(x/y - 1)/(x/y+1)*sinA

(tanB/tanC - 1)/(tanB/tanC + 1)*sinA

(tanB - tanC)/(tanB + tanC)*sinA

now multiply top and bottom by cosB.cosC,

(sinB.cosC - sinC.cosB)/(sinB.cosC + sinC.cosB)*sin(B+C)

sin(B-C)/sin(B+C)*sin(B+C)

= sin(B-C)