sin α = − 3/5 sec β = 5/4 for a third-quadrant angle α & a 1st-quadrant angle β, find the following. (A) sin(α + β) (A) tan(α + β)?

For the third-quadrant angle α :
sinα = -3/5
cosα = -√[1 - sin²α] = -√[1 - (-3/5)²] = -4/5
tanα = (-3/5) / (-4/5) = 3/4

For the third-quadrant angle β :
secβ = 5/4
cosβ = 4/5
sinα = √[1 - cos²α] = √[1 - (4/5)²] = 3/5
tanα = (3/5) / (4/5) = 3/4

(A)
sin(α + β)
= sinα cosβ + sinβ cosα
= (-3/5) × (4/5) + (3/5) × (-4/5)
= -12/25 - 12/25
= -24/25

(B)
tan(α + β)
= [tanα + tanβ] / [1 - tanα tanβ]
= [(3/4) + (3/4)] / [1 - (3/4) × (3/4)]
= (3/2) / [1 - (9/16)]
= (3/2) / (7/16)
= (3/2) × (16/7)
= 24/7

(C)
Since in (A) sin(α + β) < 0 and in (B) tan(α + β) > 0, then (α + β) is in quadrant III.

a is in (180,270), b is in (0,90). sina= (-3/5), cosa=-(1- 9/25)^(1/2)= (-4/5),
cosb = (4/5), sinb = (1 - 16/25)^(1/2) = (3/5). (A): sin(a+b) = sinacosb +
cosasinb=(-3/5)(4/5)+(-4/5)(3/5)=(-24/25). cos(a+b)= cosacosb - sinasinb
(-4/5)(4/5)-(-3/5)(3/5)=(9-16)/25=(-7/25). (B): tan(a+b)=sin(a+b)/cos(a+b),
ie., tan(a+b) = (-24/25)/(-7/25) = (24/7).

Questions A and B perhaps ?
A
sin (α + β) = sin α cos β + cos α sin β
sin (α + β) = (-3/5) (4/5) + (-4/5) (3/5) = - 24/25

B
tan (α + β) = [ tan α + tan β ] / [ 1 - tan α tan β ]
tan (α + β) = [ 3/4 + 3/4 ] / [ 1 - 9/16 ]
tan (α + β) = 24 / 7

a)
sin α = -3/5
opp/hyp = 3/5
Ignore the signs for now

opp = 3
hyp = 5
adj = sqrt(5^2 - 3^2 ) = sqrt(25-9)=sqrt(16) = 4

cos α = adj / hyp = -4/5
sin α = -3/5

sin α < 0, cos α < 0 in quadrant III

sec β = 5/4
cos β = 4/5
adj / hyp = 4/5
opp = sqrt(5^2-4^2) = sqrt(25-16) = sqrt(9) = 3

cos β = 4/5 > 0 in quadrant I
sin β = opp/hyp = 3/5 > 0 in quadrant I

sin α = -3/5
cos α = -4/5
sin β = 3/5
cos β = 4/5

sin(α+β) = sin(α)cos(β) + cos(α)sin(β)
= (-3/5)(4/5) + (-4/5)(3/5)
= -12 /25 -12/25
= -24/25

b)

cos(α+β) = cos(α)cos(β)-sin(α)sin(β)
= (-4/5)(4/5) - (-3/5)(3/5)
= -16/25 + 9/25 = -7/25

tan(α+β) = sin(α+β) / cos(α+β) = [-24/25] / [ -7/25] = 24/7
sin(α+β) < 0 and cos(α+β) < 0
so the angle α+β is in quadrant III

sin α = -3/5, and 180 < α < 270°
So cos α < 0; so cos α = -4/5 (You can use Pythagoras to check that: 3, 4, 5 triangle.)
sec β = 5/4,
so cos β = 4/5; and 0 < β < 90°
So sin β = 3/5
sin (α + β) = sin α cos β + cos α sin β
= -3/5 * 4/5 + (-4/5) * 3/5
= -24/25
tan α = sin α / cos α = 3/4
tan β = 3/4
so tan (α + β) = (tan α + tan β) / (1 - tan α tan β)
= 3/2 / (1 - 9/16)
= 3/2 / (7/16)
= 24/7
(Or first find cos (α + β), and divide.)

I know it somehow ends up in quadurant III based on the sample video but I just don't understand how to solve this question.

Tan^2 - 3/5+ cos^2