Answer:
See below
Step-by-step explanation:
We want to prove that
![\sin(x)\tan(x) = \dfrac{1}{\cos(x)} - \cos(x), \forall x \in\mathbb{R}](https://tex.z-dn.net/?f=%5Csin%28x%29%5Ctan%28x%29%20%3D%20%5Cdfrac%7B1%7D%7B%5Ccos%28x%29%7D%20-%20%5Ccos%28x%29%2C%20%5Cforall%20x%20%5Cin%5Cmathbb%7BR%7D)
Taking the RHS, note
![\dfrac{1}{\cos(x)} - \cos(x) = \dfrac{1}{\cos(x)} - \dfrac{\cos(x) \cos(x)}{\cos(x)} = \dfrac{1-\cos^2(x)}{\cos(x)}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B%5Ccos%28x%29%7D%20-%20%5Ccos%28x%29%20%3D%20%5Cdfrac%7B1%7D%7B%5Ccos%28x%29%7D%20-%20%5Cdfrac%7B%5Ccos%28x%29%20%5Ccos%28x%29%7D%7B%5Ccos%28x%29%7D%20%3D%20%5Cdfrac%7B1-%5Ccos%5E2%28x%29%7D%7B%5Ccos%28x%29%7D)
Remember that
![\sin^2(x) + \cos^2(x) =1 \implies 1- \cos^2(x) =\sin^2(x)](https://tex.z-dn.net/?f=%5Csin%5E2%28x%29%20%2B%20%5Ccos%5E2%28x%29%20%3D1%20%5Cimplies%201-%20%5Ccos%5E2%28x%29%20%3D%5Csin%5E2%28x%29)
Therefore,
![\dfrac{1-\cos^2(x)}{\cos(x)} = \dfrac{\sin^2(x)}{\cos(x)} = \dfrac{\sin(x)\sin(x)}{\cos(x)}](https://tex.z-dn.net/?f=%5Cdfrac%7B1-%5Ccos%5E2%28x%29%7D%7B%5Ccos%28x%29%7D%20%3D%20%5Cdfrac%7B%5Csin%5E2%28x%29%7D%7B%5Ccos%28x%29%7D%20%3D%20%5Cdfrac%7B%5Csin%28x%29%5Csin%28x%29%7D%7B%5Ccos%28x%29%7D)
Once
![\dfrac{\sin(x)}{\cos(x)} = \tan(x)](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csin%28x%29%7D%7B%5Ccos%28x%29%7D%20%3D%20%5Ctan%28x%29)
Then,
![\dfrac{\sin(x)\sin(x)}{\cos(x)} = \sin(x)\tan(x)](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csin%28x%29%5Csin%28x%29%7D%7B%5Ccos%28x%29%7D%20%3D%20%5Csin%28x%29%5Ctan%28x%29)
Hence, it is proved
A=7 b =4 and c=9 Because the 7 is the base of the number
(-1,2.5), (0,5), (1,10), (2,20)
Answer:
.4 in or 2/5 in per month (.4 and 2/5 means the same one is in decimal and the other one is in fractions.