Math, asked by baralsadikshya2007, 4 days ago

prove: sin(A + B) ÷ sin(A - B) = (tanA + tanB )÷ (tanA - tanB)​

Answers

Answered by jitendarm782
4

Answer:

I hope answer is helpful

Attachments:
Answered by chemist4
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)

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