I hope this helps you
✔cos^2A+sin^2A=1
✔1-cos^2A=sin^2A
✔cos2A=cos^2A-sin^2A
✔sin2A=2.sinA.cosA
secA=1/cosA
tgA=sinA/cosA
sin^2A/1/cos^2A-sin^2A/cos^2A
sin^2A.cos^2A/cos2A
2.sin^2A.cos^2A/cos2A
sin2A.2.sin2A/cos2A
tg2A.2.sin2A
Answer:
D.
Step-by-step explanation:
Remember that the limit definition of a derivative at a point is:
![\displaystyle{\frac{d}{dx}[f(a)]= \lim_{x \to a}\frac{f(x)-f(a)}{x-a}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28a%29%5D%3D%20%5Clim_%7Bx%20%5Cto%20a%7D%5Cfrac%7Bf%28x%29-f%28a%29%7D%7Bx-a%7D%7D)
Hence, if we let f(x) be ln(x+1) and a be 1, this will yield:
![\displaystyle{\frac{d}{dx}[f(1)]= \lim_{x \to 1}\frac{\ln(x+1)-\ln(2)}{x-1}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%281%29%5D%3D%20%5Clim_%7Bx%20%5Cto%201%7D%5Cfrac%7B%5Cln%28x%2B1%29-%5Cln%282%29%7D%7Bx-1%7D%7D)
Hence, the limit is equivalent to the derivative of f(x) at x=1, or f’(1).
The answer will thus be D.
Answer:
A
C
i don't get the first slide
Step-by-step explanation:
This is a backup question it would be great to state what question 3 was please
*written: seven million.
*expanded: 7,000,000+000,000