Solution:

Given: sin ( A B) =1 and sin ( A-B )=1/2

We are supposed to find the value of tan (A 2B) and tan (2A B).

Approach: We will use the trigonometric formulae to solve the problem.

Trigonometric Formulae:

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

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

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

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

Calculation:

Finding value of sin A and sin B:

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

Put A = B in the above equation to get:

sin (2A) = 2 sin A cos A

=> sin A = (1/2)

Similarly, using sin (A-B) = sin A cos B - cos A sin B and putting A = 2B and B = A/2, we get:

sin (3B / 2) = (sqrt(3) / 2) sin B

=> sin B = (sqrt(3) / 3)

Finding value of cos A and cos B:

Using sin^2 A + cos^2 A = 1, we get:

cos A = (sqrt(3) / 2)

Similarly, using sin^2 B + cos^2 B = 1, we get:

cos B = (1/2)

Finding value of tan (A 2B) and tan (2A B):

Using the formula, tan (A+B) = (tan A + tan B) / (1 - tan A tan B), we get:

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

=> tan (A 2B) = (2sqrt(3)) / 3

Similarly, using the formula, tan (A-B) = (tan A - tan B) / (1 + tan A tan B), we get:

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

=> tan (2A B) = (sqrt(3)) / 3

Answer:

The value of tan (A 2B) is (2sqrt(3)) / 3 and the value of tan (2A B) is (sqrt(3)) / 3.