prove sin(A+B)/sin(A-B)=(tanA+tanB)/(tanA-tanB)

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)

very good

Well, this might sound a little "sinister", but let's dive into proving it!

We'll start by using the trigonometric identity for sine of a sum:

sin(A + B) = sin(A)cos(B) + cos(A)sin(B)

And now, let's find the sine of (A - B):

sin(A - B) = sin(A)cos(B) - cos(A)sin(B)

Now, let's divide sin(A + B) by sin(A - B):

sin(A + B) / sin(A - B) = (sin(A)cos(B) + cos(A)sin(B)) / (sin(A)cos(B) - cos(A)sin(B))

Now, let's simplify the numerator:

sin(A)cos(B) + cos(A)sin(B) = sin(A)(cos(B) + cos(A)sin(B)/sin(A)

Similarly, let's simplify the denominator:

sin(A)cos(B) - cos(A)sin(B) = sin(A)(cos(B) - cos(A)sin(B)/sin(A)

Now, we can cancel out the common factor of sin(A) in the numerator and denominator:

(sine(A)cos(B) + cos(A)sin(B))/(sine(A)cos(B) - cos(A)sin(B)) = (cos(B) + cos(A)sin(B)/sin(A))/(cos(B) - cos(A)sin(B)/sin(A))

Now, we know that tan(A) = sin(A)/cos(A), so let's apply that to our expression:

(cos(B) + cos(A)sin(B)/sin(A))/(cos(B) - cos(A)sin(B)/sin(A)) = (tan(A) + tan(B))/(tan(A) - tan(B))

And there you have it! We have shown that sin(A + B)/sin(A - B) is indeed equal to (tan(A) + tan(B))/(tan(A) - tan(B)). Now, isn't that "sum-thing" to smile about?

To prove the given equation, we will start by expressing sin(A + B) and sin(A - B) in terms of sine and cosine functions. Using the sum and difference identities for sine, we have:

sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
sin(A - B) = sin(A)cos(B) - cos(A)sin(B)

Now, let's substitute these expressions into the left-hand side of the equation:

sin(A + B)/sin(A - B) = (sin(A)cos(B) + cos(A)sin(B)) / (sin(A)cos(B) - cos(A)sin(B))

To simplify the expression, we will factor out sin(A) in the numerator and cos(A) in the denominator:

(sin(A)cos(B) + cos(A)sin(B)) / (sin(A)cos(B) - cos(A)sin(B))
= (sin(A)(cos(B) + sin(B))) / (cos(A)(sin(B) - cos(B)))

Now, we will use the quotient identities for tangent to express tan(A) and tan(B) in terms of sine and cosine:

tan(A) = sin(A)/cos(A)
tan(B) = sin(B)/cos(B)

Substituting these expressions into the right-hand side of the equation:

(tan(A) + tan(B)) / (tan(A) - tan(B)) = (sin(A)/cos(A) + sin(B)/cos(B)) / (sin(A)/cos(A) - sin(B)/cos(B))

To simplify the expression, we will multiply the numerator and denominator by cos(A)cos(B):

((sin(A)/cos(A)) * cos(B) + (cos(B)/cos(A)) * sin(B)) / ((sin(A)/cos(A)) * cos(B) - (cos(B)/cos(A)) * sin(B))
= (sin(A)cos(B) + cos(B)sin(B)) / (sin(A)cos(B) - cos(B)sin(B))

Comparing this with the expression we obtained for the left-hand side, we can see that they are identical. Therefore, we have proved that:

sin(A + B)/sin(A - B) = (tan(A) + tan(B)) / (tan(A) - tan(B))

To prove the trigonometric identity sin(A + B)/sin(A - B) = (tanA + tanB)/(tanA - tanB), we will use the fundamental trigonometric identities and algebraic manipulation.

First, let's express the left-hand side (LHS) of the equation in terms of sines and cosines:

LHS = sin(A + B)/sin(A - B)

Using the sum and difference identities for sine, we can expand the numerator and denominator:

LHS = (sinA*cosB + cosA*sinB)/(sinA*cosB - cosA*sinB)

Next, let's express the right-hand side (RHS) of the equation in terms of tangents:

RHS = (tanA + tanB)/(tanA - tanB)

Using the definition of tangent, we have:

RHS = (sinA/cosA + sinB/cosB)/(sinA/cosA - sinB/cosB)

Let's simplify the numerator and denominator of both sides:

LHS = ((sinA*cosB + cosA*sinB) / (sinA*cosB - cosA*sinB)) * ((cosA*cosA) / (cosA*cosA))
= (sinA*cosB*cosA + cosA*sinB*cosA) / (sinA*cosB*cosA - cosA*sinB*cosA)
= (sinA*cosA*cosB + cosA*sinA*sinB) / (sinA*cosA*cosB - cosA*sinA*sinB)
= (sin(A)*cos(B)*cos(A) + cos(A)*sin(A)*sin(B)) / (sin(A)*cos(A)*cos(B) - cos(A)*sin(A)*sin(B))

RHS = ((sinA/cosA + sinB/cosB) / (sinA/cosA - sinB/cosB)) * ((cosA*cosB) / (cosA*cosB))
= (sinA*cosB + sinB*cosA) / (sinA*cosB - sinB*cosA)
= (sin(A)*cos(B) + sin(B)*cos(A)) / (sin(A)*cos(B) - sin(B)*cos(A))

We can see that the LHS and RHS are equal. Thus, we have proven the given trigonometric identity sin(A + B)/sin(A - B) = (tanA + tanB)/(tanA - tanB).