Verified answer

To prove that tan(A) = (1 - tan(B))/(1 + tan(B)), given sin(A + B) = cos(A + B), we can use the following trigonometric identities:

1. sin(A + B) = cos(A + B) can be rewritten as:

  sin(A + B) - cos(A + B) = 0

2. Using the sum-to-product identity for sine and cosine:

  sin(A + B) - cos(A + B) = 0

  ⇒ sin(A)cos(B) + cos(A)sin(B) - cos(A)cos(B) + sin(A)sin(B) = 0

3. Rearranging the terms:

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

4. Applying the difference-to-product identity for cosine:

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

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

5. Using the identity sin(-x) = -sin(x):

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

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

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

6. Factoring out sin(A) and cos(B) on the left side, and sin(B) and cos(A) on the right side:

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

7. Dividing both sides by cos(B) - sin(B) and cos(A) - sin(A):

  sin(A) = sin(B)

  ⇒ A = B + kπ, where k is an integer

8. Now, let's calculate tan(A) and tan(B):

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

9. Substituting tan(A) = tan(B) into the equation we need to prove:

  (1 - tan(B))/(1 + tan(B)) = (1 - tan(B))/(1 + tan(B))  (identity)

Hence, we have proved that if sin(A + B) = cos(A + B), then tan(A) = (1 - tan(B))/(1 + tan(B)).

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Answer:

To prove that tan(A) = (1 - tan(B))/(1 + tan(B)) using the given condition sin(A+B) = cos(A+B), we can start by expressing sin(A+B) and cos(A+B) in terms of tan(A) and tan(B).

Let's begin by expanding sin(A+B) and cos(A+B) using the identities for sine and cosine:

sin(A+B) = sin(A)cos(B) + cos(A)sin(B) (1)

cos(A+B) = cos(A)cos(B) - sin(A)sin(B) (2)

Now, we can square both sides of the given equation sin(A+B) = cos(A+B):

(sin(A)cos(B) + cos(A)sin(B))^2 = (cos(A)cos(B) - sin(A)sin(B))^2

Expanding both sides:

sin^2(A)cos^2(B) + 2sin(A)cos(A)sin(B)cos(B) + cos^2(A)sin^2(B) = cos^2(A)cos^2(B) - 2sin(A)cos(A)sin(B) + sin^2(A)sin^2(B)

Next, we simplify and cancel out the common terms on both sides:

sin^2(A)cos^2(B) + cos^2(A)sin^2(B) = cos^2(A)cos^2(B) + sin^2(A)sin^2(B)

Rearranging the terms:

sin^2(A)cos^2(B) - sin^2(A)sin^2(B) = cos^2(A)cos^2(B) - cos^2(A)sin^2(B)

Factoring out sin^2(A) and cos^2(B) on the left side, and cos^2(A) and sin^2(B) on the right side:

sin^2(A)(cos^2(B) - sin^2(B)) = cos^2(A)(cos^2(B) - sin^2(B))

Using the identity sin^2(x) + cos^2(x) = 1, we can simplify the equation:

sin^2(A) = cos^2(A)

Now, divide both sides by cos^2(A):

sin^2(A)/cos^2(A) = cos^2(A)/cos^2(A)

This simplifies to:

tan^2(A) = 1

Taking the square root of both sides:

tan(A) = ±1

Now, consider the equation we want to prove:

tan(A) = (1 - tan(B))/(1 + tan(B))

We can rewrite the right side as:

tan(A) = (1 - tan(B))/(1 + tan(B)) * (tan(B)/tan(B))

Simplifying:

tan(A) = (tan(B) - tan^2(B))/tan(B) = tan(B)(1 - tan(B))/tan(B) = 1 - tan(B)

Therefore, we have shown that tan(A) = 1 - tan(B).