Remember: We have to work from either the LHS or the RHS.
(Left hand side or the Right hand side)
You should already know this:
![\huge{Cot(t) = \frac{1}{tan(t)} = \frac{1}{\frac{sin(t)}{cos(t)}} = 1\div \frac{sin(t)}{cos(t)} = 1\times \frac{cos(t)}{sin(t)}=\boxed{\frac{cos(t)}{sin(t)}}](https://tex.z-dn.net/?f=%5Chuge%7BCot%28t%29%20%3D%20%5Cfrac%7B1%7D%7Btan%28t%29%7D%20%3D%20%5Cfrac%7B1%7D%7B%5Cfrac%7Bsin%28t%29%7D%7Bcos%28t%29%7D%7D%20%3D%201%5Cdiv%20%5Cfrac%7Bsin%28t%29%7D%7Bcos%28t%29%7D%20%3D%201%5Ctimes%20%5Cfrac%7Bcos%28t%29%7D%7Bsin%28t%29%7D%3D%5Cboxed%7B%5Cfrac%7Bcos%28t%29%7D%7Bsin%28t%29%7D%7D)
You should also know this:
![sin^2(t) + cos^2(t) = 1\\\\\boxed{sin^2(t)} = 1 - cos^2(t)](https://tex.z-dn.net/?f=sin%5E2%28t%29%20%2B%20cos%5E2%28t%29%20%3D%201%5C%5C%5C%5C%5Cboxed%7Bsin%5E2%28t%29%7D%20%3D%201%20-%20cos%5E2%28t%29)
So plugging in both of those into our identity, we get:
![\frac{cos(t)}{sin(t)}\cdot sin^2(t) = cos(t)\cdot sin(t)](https://tex.z-dn.net/?f=%5Cfrac%7Bcos%28t%29%7D%7Bsin%28t%29%7D%5Ccdot%20sin%5E2%28t%29%20%3D%20cos%28t%29%5Ccdot%20sin%28t%29)
Simplify the denominator on the LHS (Left Hand Side)
We get:
![cos(t) \cdot sin(t) = cos(t) \cdot sin(t)](https://tex.z-dn.net/?f=cos%28t%29%20%5Ccdot%20sin%28t%29%20%3D%20cos%28t%29%20%5Ccdot%20sin%28t%29)
LHS = RHS
Therefore, identity is verified.
Answer:
here u go
Step-by-step explanation:
it would be 0.075,3/4, 75%
Pretty sure it’s C. hope this helps
Answer:
If there are 10 students taking only chemistry, 9 students taking only physics, and 5 students only taking both chemisty and 16 students are taking neither; I would add 10+9+5+16=40 (total students) and divide 10/40 (25% chemistry) 9/40 (22.5% physics) 5/40 (12.5% both) 16/40 (40% neither)
Step-by-step explanation:
1. Determine a single event with a single outcome.
2. Identify the total number of outcomes that can occur.
3. Divide the number of events by the number of possible outcomes.
Answer:
B) (7x + 6y)(7x − 6y)
Step-by-step explanation: