Sin A=-4/5 Sec B =5/3. A is a third quadrant angle B is a first quadrant angle What are sin(A+B) and tan(A+B)?

Answer 1

sin A = - 4 / 5
cos B = 3 / 5

sin (A + B) = sin A cos B + cos A sin B
sin (A + B) = (- 4/5) (3/5) + (- 3/5)(4/5)
sin (A + B) = (- 24 / 25)

tan(A + B) = (tan A + tan B) / (1 - tan A tan B)
tan(A + B) = (4/3 + 4/3) / (1 - 16/9)
tan(A + B) = (8/3) / (- 7 / 9)
tan (A + B) = - 72 / 21 = - 24 / 7

Answer 2

This is not quite a trick question, but you should recognize that you're dealing with 3/4/5 right triangles. A is appx 270-36 deg or 234 deg. B is about 54 degrees (cosine B=3/5).
Use the trig identities to find sin(A+B) and tan(A+B). For example, sin(A+B)= sinAcosB+cosAsinB. We know sinA and cosB. CosA=-3/5 and sinB=4/5. The corresponding identify for cos(A+B) is CosACosB-SinASinB.
Then tan(A+B)= sin(A+B)/cos(A+B)

Answer 3

sinA = -4/5
sin²A+cos²A = 1
cos²A = 1 - 16/25 = 9/25
A is in the third quadrant
cosA is negative
cosA = -3/5

sinA = -4/5
cosA = -3/5

secB = 5/3
cosB = 1/secB = 3/5
B is in the first quadrant
sin²B+cos²B = 1

sinB = 4/5
cosB = 3/5

sin(A+B)
= sinAcosB + cosAsinB
= (-4/5)(3/5) + (-3/5)(4/5)
= -12/25 - 12/25
= -24/25

cos(A+B)
= cosAcosB - sinAsinB
= (-3/5)(3/5) - (-4/5)(4/5)
= -9/25 + 16/25
= 7/25

tan(A+B)
= sin(A+B)/cos(A+B)
= (-24/25)/(7/25)
= -24/7


sin(A+B) = -24/25
tan(A+B) = -24/7