Proof:

We are given that tan(a) = k tan(b).

To prove: sin(ab) = (k/(k-1)) sin(a-b)

Step 1: Expressing tan(a) and tan(b) in terms of sin and cos:

Recall that the tangent function is defined as the ratio of the sine and cosine functions:

tan(x) = sin(x) / cos(x)

Therefore, we can rewrite the given equation as:

sin(a) / cos(a) = k * (sin(b) / cos(b))

Step 2: Cross-multiplying and simplifying:

Multiplying both sides of the equation by cos(a) * cos(b), we have:

sin(a) * cos(b) = k * sin(b) * cos(a)

Using the trigonometric identity sin(a) * cos(b) = (1/2) * [sin(a+b) + sin(a-b)], we can rewrite the left side of the equation as:

(1/2) * [sin(a+b) + sin(a-b)] = k * sin(b) * cos(a)

Step 3: Simplifying further:

Expanding the right side of the equation, we have:

(1/2) * [sin(a+b) + sin(a-b)] = (1/2) * [2 * k * sin(b) * cos(a)]

Canceling out the common factors, we get:

sin(a+b) + sin(a-b) = 2 * k * sin(b) * cos(a)

Step 4: Applying the trigonometric identity sin(a+b) = sin(a) * cos(b) + cos(a) * sin(b) and sin(a-b) = sin(a) * cos(b) - cos(a) * sin(b), we can rewrite the equation as:

[sin(a) * cos(b) + cos(a) * sin(b)] + [sin(a) * cos(b) - cos(a) * sin(b)] = 2 * k * sin(b) * cos(a)

Simplifying further, we have:

2 * sin(a) * cos(b) = 2 * k * sin(b) * cos(a)

Step 5: Canceling out the common factor of 2, we get:

sin(a) * cos(b) = k * sin(b) * cos(a)

Step 6: Rearranging the terms, we have:

sin(a) / sin(b) = k * cos(a) / cos(b)

Step 7: Using the trigonometric identity sin(x) = √(1 - cos^2(x)), we can rewrite the equation as:

√(1 - cos^2(a)) / √(1 - cos^2(b)) = k * cos(a) / cos(b)

Squaring both sides of the equation, we get:

(1 - cos^2(a)) / (1 - cos^2(b)) = (k^2 * cos^2(a)) / (cos^2(b))

Step 8: Simplifying the equation further:

Expanding the numerator and denominator, we have:

1 - cos^2(a) = k^2 * cos^2(a) * (1 - cos^2(b))

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