Proof:

Given:
A * B = x (equation 1)
sin A = k * sin B (equation 2)

To prove:
tan A = k * sin(x) / (1 - cos(x))

Proof:

Using the identity sin^2(A) + cos^2(A) = 1, we can rewrite equation 2 as follows:

sin^2(A) = k^2 * sin^2(B)
1 - cos^2(A) = k^2 * (1 - cos^2(B))
cos^2(A) = 1 - k^2 + k^2 * cos^2(B) (equation 3)

Now, let's express cos^2(B) in terms of cos(x) and sin(x) using the identity cos^2(B) = 1 - sin^2(B):

cos^2(B) = 1 - sin^2(B)
cos^2(B) = 1 - (sin^2(x) / cos^2(x))
cos^2(B) = (cos^2(x) - sin^2(x)) / cos^2(x)
cos^2(B) = (1 - sin^2(x)) / cos^2(x)
cos^2(B) = cos^2(x) / cos^2(x) (equation 4)

Substituting equation 4 into equation 3, we get:

cos^2(A) = 1 - k^2 + k^2 * (cos^2(x) / cos^2(x))
cos^2(A) = 1 - k^2 + k^2
cos^2(A) = 1 (equation 5)

Taking the square root of equation 5, we have:

cos(A) = 1

Since cos(A) cannot be equal to 1 (as it would imply A = 0), there must be a mistake in the given equations. Please double-check the original problem statement and equations for any errors.

Just replace,