An agale whose measure is 40 is in standard positition. In which quadrant does the terminal side of the angle fall? it would be the 1st Quadrant
Answer:
Answer my question
Step-by-step explanation:
15is the correct answer
If we take the Pythagorean identity identity sin^2 x + cos^2 x = 1 then
<span>(cos^2 x + sin^2 x) / (cot^2 x - csc^2 x)
The numerator becomes 1 since addition order matters not.
1 / </span>(cot^2 x - csc^2 x)
If we factor the denominator out a negative
1 / -(<span>csc^2 x - cot^2 x)
Consider </span><span>sin^2 x + cos^2 x = 1. Divide both sides by sin^2 x to get
1 + cot^2 x = csc^2 x
Subtract both sides by cot^2 x to get 1 = csc^2 x - cot^2 x.
Replace the denominator
1 / -(1) = -1
For cos</span>^2 θ / sin^2 θ + csc θ sin θ, we use cscθ = 1/sinθ and cosθ/sinθ = cotθ so
= cos^2 θ / sin^2 θ + 1
= cot^2 θ + 1
We use 1 + cot^2 <span>θ = csc^2 </span>θ to simplify this to
= csc^2 θ
Answers: -1
csc^2 θ
If this is a triangle the the measure of angles 11 and 6 could be any number between 1 and 60
Answer:
3m^3/8 + 3
Step-by-step explanation:
It can't be solved but that's the most simplified version.
