anonymous one year ago Verify the identity. Show your work. 1 + sec2xsin2x = sec2x the answer is 2

1. anonymous

Teacher:Substitute for the sec^2x on the RHS only. This is a reciprocal function. You should be able to take this from here. Me: wtf?

2. anonymous

@thomaster

3. anonymous

@Etrainx

4. anonymous

@Zale101

5. anonymous

6. anonymous

i used that one but my teacher said that up there so im confuzled

7. anonymous

@mukushla

8. anonymous

@zepdrix

9. anonymous

@poopsiedoodle

10. anonymous

@mathstudent55

11. anonymous

@TheSmartOne

12. anonymous

@Vocaloid

13. anonymous

@Vocaloid can you give me some understanding of my teachers words plz

14. Vocaloid

sec^2(x) = 1/(cos^2(x))

15. Vocaloid

that's basically what she's saying, substitute 1/(cos^2(x)) for sec^2(x) on the right side

16. anonymous

|dw:1439233166138:dw|

17. anonymous

this is to confusing this is why i hate math lol

18. anonymous

That's why I'm helping you. :D

19. anonymous

Do you know how to multiply this?

20. anonymous

true that thnx alot lol

21. anonymous

i really dont

22. anonymous

For instance, what is $$4\times \frac{ 1 }{ 3}=?$$ You only need to put 4 in the numerator. So you'll get $$\frac{ 4 }{ 3 }$$.

23. anonymous

Now, what is $$1 + (\frac{ 1 }{ \cos^2x }\times \sin^2x)$$?

24. anonymous

hold on ups is at my gate

25. anonymous

imma go see what it is

26. anonymous

imm home alone so i have to get it smh lazy people deez days i leave my gate open for a reason haha

27. anonymous

@mathway im back now

28. anonymous

29. anonymous

frac(cos^2(x))

30. anonymous

is that right @mathway

31. anonymous

??

32. misty1212

HI!!

33. misty1212

it is clear that $\frac{\sin(x)}{\cos(x)}=\tan(x)$?

34. anonymous

hiii

35. anonymous

yes

36. misty1212

and also it is clear that $\sec(x)=\frac{1}{\cos(x)}$right?

37. anonymous

yess

38. misty1212

that means $\sec^2(x)\sin^2(x)=\frac{\sin^2(x)}{\cos^2(x)}=\tan^2(x)$

39. anonymous

ok

40. misty1212

so you are looking at $1+\tan^2(x)=\sec^2(x)$ which is definitely true, we can get it in one step

41. misty1212

start with the mother of all trig identities $\cos^2(x)+\sin^2(x)=1$ divide both sides by $$\cos^2(x)$$ and you get $\frac{\cos^2(x)}{\cos^2(x)}+\frac{\sin^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}$ or $1+\tan^2(x)=\sec^2(x)$ as needed

42. anonymous

woow thats alot to take in lol but it makes since

43. misty1212

$\color\magenta\heartsuit$