TRIGONOMETRY

[r267747]

Ques:Sin(A+B)/Cos(A-B)=1-m/1+m.
Show that tan(pi/4-A).tan(pi-B)=m
MY WORK:
Apply commodento & dividedo
Sin(A+B)+Cos(A-B)/Sin(A+B)-Cos(A-B)=1-m+1+m/1-m-1-m
Sin(A+B)+Sin(pi/2-(A-B))/Sin(A+B)-Sin(pi/2-(A-B))=-1/m
Sin(pi/4+B).Cos(-pi/4+A)/Cos(pi/4+B).Sin(-pi/4+A)=-1/m
tan(pi/4+B).Cot(-pi/4+B)=-1/m
SIR,NOW I DON'T KNOW WHAT TO DO.
SO,PLEASE SOLVE IT

 

[r267747]

Ques:Sin(A+B)/Cos(A-B)=1-m/1+m.
Show that tan(pi/4-A).tan(pi-B)=m
MY WORK:
Apply commodento & dividedo
Sin(A+B)+Cos(A-B)/Sin(A+B)-Cos(A-B)=1-m+1+m/1-m-1-m
Sin(A+B)+Sin(pi/2-(A-B))/Sin(A+B)-Sin(pi/2-(A-B))=-1/m
use formula (SinA+SinB),Sin(pi/4+B).Cos(-pi/4+A)/Cos(pi/4+B).Sin(-pi/4+A)=-1/m
tan(pi/4+B).Cot(-pi/4+B)=-1/m
SIR,NOW I DON'T KNOW WHAT TO DO.
SO,PLEASE SOLVE IT

 

[Deleted member 4993]

Ques:Sin(A+B)/Cos(A-B)=1-m/1+m.
Show that tan(pi/4-A).tan(pi-B)=m
MY WORK:
Apply commodento & dividedo
Sin(A+B)+Cos(A-B)/Sin(A+B)-Cos(A-B)=1-m+1+m/1-m-1-m
Sin(A+B)+Sin(pi/2-(A-B))/Sin(A+B)-Sin(pi/2-(A-B))=-1/m
use formula (SinA+SinB),Sin(pi/4+B).Cos(-pi/4+A)/Cos(pi/4+B).Sin(-pi/4+A)=-1/m
tan(pi/4+B).Cot(-pi/4+B)=-1/m
SIR,NOW I DON'T KNOW WHAT TO DO.
SO,PLEASE SOLVE IT

\(\displaystyle \frac{cos(A-B) - sin(A+B)}{cos(A-B) + sin(A+B)} = m\)

\(\displaystyle \frac{1 + tan(A)*tan(B) - tan(A) - tan(B)}{1 + tan(A)*tan(B) - tan(A) - tan(B)} = m\)

\(\displaystyle \frac{[1 - tan(A)]*[1-tan(B)]}{[1 + tan(A)]*[1+tan(B)]} = m\)

Now continue....