Find tan(alpha-beta)

Question

Sin alpha= 4/5 with alpha in quadrant I

tan beta= 3/4 with beta in quadrant III

Answer 1

sin(α) = 4/5 (in Q1):

Using the Pythagorean Theorem, x = √(5² - 4²) = √9 = 3

So tan(α) = 4/3

Using the sum/difference identities and the fact that tan(β) = 3/4:

tan(α - β) = [tan(α) - tan(β)]/[1 + tan(α)tan(β)]

tan(α - β) = (4/3 - 3/4)/[1 + (4/3)(3/4)]

tan(α - β) = (16/12 - 9/12)/(1 + 1)

tan(α - β) = (7/12)/2

tan(α - β) = 7/24

Answer 2

sin A = 4/5

cos A = 3/5

tan A = 4/3

tan B = 3/4

sin B = 3/5 = cos A

cos B = 4/5 = sin A

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

= (4/3 - 3/4) / (1 + 4/3*3/4)

= (4/3 - 3/4) / (1 + 1)

= (1/2) ( 4/3 - 3/4)

= (1/2) (12/12 - 9/12)

= (1/2) ( 7/12)

= 7/24