prove: sin(A + B) ÷ sin(A - B) = (tanA + tanB )÷ (tanA - tanB)
Answers
Answered by
4
Answer:
I hope answer is helpful
Attachments:
Answered by
0
Answer:
tan(A+B) = (tanA+tanB)/(1-tanAtanB)
tan(A-B) = (tanA-tanB)/(1+tanAtanB)
so,
tanA+tanB = tan(A+B)(1-tanAtanB)
tanA-tanB = tan(A-B)(1+tanAtanB)
divide the two and you have
tan(A+B)/tan(A-B) * (1-tanAtanB)/(1+tanAtanB)
multiply top and bottom by cosAcosB nad you have
tan(A+B)/tan(A-B) * (cosAcosB-sinAsinB)/(cosAcosB+sinAsinB)
That's just
tan(A+B)/tan(A-B) * cos(A+B)/cos(A-B)
= sin(A+B)/sin(A-B)
Similar questions