If 5cosA+3=0 and 180(degrees)<A<360(degrees)

Determine

1. Tan^2A
And
2. SinA over cosA

cosA = -3/5 so A is ins QIII

sinA = -4/5

now the rest is easy...

Hhgcbff

To determine the value of (1) Tan^2A and (2) SinA over cosA, we need to first find the value of A using the given equation 5cosA + 3 = 0.

Step 1: Solve the equation 5cosA + 3 = 0 for cosA.
Subtract 3 from both sides: 5cosA = -3
Divide both sides by 5: cosA = -3/5

Step 2: Use the inverse cosine function (cos^(-1)) to find the value of A.
Take the inverse cosine of both sides: A = cos^(-1)(-3/5)

Step 3: Evaluate cos^(-1)(-3/5) to find the value of A.
Use a calculator or a trigonometric table to find the value of cos^(-1)(-3/5).
The result is A ≈ 143.13 degrees.

Now, we can calculate (1) Tan^2A and (2) SinA over cosA using the value of A.

1. Tan^2A:
Recall that Tan^2A is equal to (SinA)^2 / (CosA)^2.
Since SinA over cosA is also involved, we can use the Pythagorean identity Sin^2A + Cos^2A = 1.

Using the given equation: 5cosA + 3 = 0
Substitute cosA = -3/5:
5(-3/5) + 3 = 0
-3 + 3 = 0
0 = 0

Since the equation simplifies to 0 = 0, this means that the equation is always true, regardless of the value of A. Therefore, Tan^2A is indeterminate in this case.

2. SinA over cosA:
Calculate the value of SinA using the equation SinA = √(1 - Cos^2A).
Substituting cosA = -3/5:
SinA = √(1 - (-3/5)^2)
≈ √(1 - 9/25)
≈ √(16/25)
≈ 4/5

Now, we can calculate SinA over cosA:
SinA / cosA = (4/5) / (-3/5)
= -4/3

Therefore:
1. Tan^2A is indeterminate.
2. SinA over cosA is equal to -4/3.