Certainly! We start with the given equation:
[tex]\[
\sin(A + B) = k \sin(A - B)
\][/tex]
To prove the relationship [tex]\((k - 1) \tan A = (k + 1) \tan B\)[/tex], we will use trigonometric identities.
First, recall the trigonometric identities for the sine of sums and differences:
[tex]\[
\sin(A + B) = \sin A \cos B + \cos A \sin B
\][/tex]
[tex]\[
\sin(A - B) = \sin A \cos B - \cos A \sin B
\][/tex]
Using these identities, we can rewrite the given equation as:
[tex]\[
\sin A \cos B + \cos A \sin B = k (\sin A \cos B - \cos A \sin B)
\][/tex]
Now, expand and distribute the [tex]\(k\)[/tex] on the right-hand side:
[tex]\[
\sin A \cos B + \cos A \sin B = k \sin A \cos B - k \cos A \sin B
\][/tex]
Next, we gather like terms involving [tex]\(\sin A \cos B\)[/tex] and [tex]\(\cos A \sin B\)[/tex] on each side:
[tex]\[
\sin A \cos B + \cos A \sin B - k \sin A \cos B + k \cos A \sin B = 0
\][/tex]
[tex]\[
\sin A \cos B (1 - k) + \cos A \sin B (1 + k) = 0
\][/tex]
Now, the equation can be divided into two separate fractions:
[tex]\[
\frac{\cos A \sin B (1 + k)}{\cos B \sin A (1 - k)} = 1
\][/tex]
By canceling out [tex]\(\cos B\)[/tex] and [tex]\(\sin A\)[/tex] (assuming they are non-zero),
[tex]\[
\frac{\cos A}{\sin A} \cdot \frac{\sin B}{\cos B} = -\frac{1 + k}{1 - k}
\][/tex]
Recognizing the respective trigonometric functions, we can rewrite it as:
[tex]\[
\tan A = - \frac{1 + k}{1 - k} \cot B
\][/tex]
Since [tex]\(\cot B = \frac{1}{\tan B}\)[/tex],
[tex]\[
\tan A = - \frac{1 + k}{1 - k} \cdot \frac{1} {\tan B} \implies \tan A \cdot (1 - k) = - (1 + k) \cdot \frac{1}{\tan B}
\][/tex]
Finally, multiplying both sides of the equation by [tex]\(\tan B\)[/tex]:
[tex]\[
\tan A \cdot (1 - k) \cdot \tan B = - (1 + k)
\][/tex]
The minus sign reverses when we rearrange,
[tex]\[(k - 1) \tan A = (k + 1) \tan B\][/tex]
Thus, we have shown that:
[tex]\[
(k-1) \tan A = (k+1) \tan B
\][/tex]
as required.
