Mrs. Bready has a large bag filled with red and green cards. She tells the class that 15% of the cards are red and 85% are green
2 answers:
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There are three 'Pythagorean' identities that we can look at and they are
sin²(x) + cos²(x) = 1
tan²(x) + 1 = sec²(x)
1 + cot²(x) = csc²(x)
We can start by checking each option to see which one would give us any of the 'Pythagorean' identities as its simplest form
Option A:
sin²(x) sec²(x) + 1 = tan²(x) csc²(x)
Rewriting sec²(x) as 1/cos²(x)
Rewriting tan²(x) as sin²(x)/cos²(x)
Rewriting csc²(x) as 1/sin²(x)
We have
![sin^{2}(x)[ \frac{1}{ cos^{2}(x) }]+1=[ \frac{ sin^{2}( x)}{ cos^{2} (x)}][ \frac{1}{ sin^{2}(x) } ]](https://tex.z-dn.net/?f=sin%5E%7B2%7D%28x%29%5B%20%5Cfrac%7B1%7D%7B%20cos%5E%7B2%7D%28x%29%20%7D%5D%2B1%3D%5B%20%5Cfrac%7B%20sin%5E%7B2%7D%28%20x%29%7D%7B%20cos%5E%7B2%7D%20%28x%29%7D%5D%5B%20%5Cfrac%7B1%7D%7B%20sin%5E%7B2%7D%28x%29%20%7D%20%5D)
![[\frac{ sin^{2}(x) }{ cos^{2}(x) } ]+1= \frac{1}{ cos^{2}(x) }](https://tex.z-dn.net/?f=%20%5B%5Cfrac%7B%20sin%5E%7B2%7D%28x%29%20%7D%7B%20cos%5E%7B2%7D%28x%29%20%7D%20%5D%2B1%3D%20%5Cfrac%7B1%7D%7B%20cos%5E%7B2%7D%28x%29%20%7D%20)
Option B:
sin²(x) - cos²(x) = 1
This expression is already in the simplest form, cannot be simplified further
Option C:
[ csc(x) + cot(x) ]² = 1
Rewriting csc(x) as 1/sin(x)
Rewriting cot(x) as cos(x)/sin(x)
We have
![[ \frac{1}{sin(x)}+ \frac{cos(x)}{sin(x)}] ^{2} =1](https://tex.z-dn.net/?f=%5B%20%5Cfrac%7B1%7D%7Bsin%28x%29%7D%2B%20%5Cfrac%7Bcos%28x%29%7D%7Bsin%28x%29%7D%5D%20%5E%7B2%7D%20%3D1)
Option D:
csc²(x) + cot²(x) = 1
Rewriting csc²(x) as 1/sin²(x) and cot²(x) as cos²(x)/sin²(x)
from our working out we can see that option A simplified into one of 'Pythagorean' identities, hence the correct answer
sin²(x) + cos²(x) = 1
tan²(x) + 1 = sec²(x)
1 + cot²(x) = csc²(x)
We can start by checking each option to see which one would give us any of the 'Pythagorean' identities as its simplest form
Option A:
sin²(x) sec²(x) + 1 = tan²(x) csc²(x)
Rewriting sec²(x) as 1/cos²(x)
Rewriting tan²(x) as sin²(x)/cos²(x)
Rewriting csc²(x) as 1/sin²(x)
We have
Option B:
sin²(x) - cos²(x) = 1
This expression is already in the simplest form, cannot be simplified further
Option C:
[ csc(x) + cot(x) ]² = 1
Rewriting csc(x) as 1/sin(x)
Rewriting cot(x) as cos(x)/sin(x)
We have
Option D:
csc²(x) + cot²(x) = 1
Rewriting csc²(x) as 1/sin²(x) and cot²(x) as cos²(x)/sin²(x)
from our working out we can see that option A simplified into one of 'Pythagorean' identities, hence the correct answer
B would be the correct answer, 