## VeroZarate Group Title Verify that cot x sec4x = cot x + 2 tan x + tan3x one year ago one year ago

1. VeroZarate

so i know cotx is cos/sin

2. VeroZarate

idk if we can change the tan though to sin/cos?

3. binarymimic

this one is hard but try this

4. binarymimic

first divide everything by cot x

5. binarymimic

$\sec^{4} x = 1 + 2\frac{ \tan x }{ \cot x } + \frac{ \tan^{3} x }{ \cot x }$

6. binarymimic

and we know that $\frac{ \tan x }{ \cot x } = \frac{ \frac{ \sin x }{ \cos x } }{ \frac{ \cos x }{ \sin x } } = \frac{\sin^{2} x }{ \cos^{2} x } = \tan^{2} x$

7. VeroZarate

would it apply to the second one but it would be (sin^3x/cos^3x)/cosx/sinx ?

8. binarymimic

that would simplify it to: $\sec^{4} x = 1 + 2 \tan^{2} x + \tan^{4} x$

9. binarymimic

yeah the second one is: $\frac{ \tan^{3} x }{ \cot x } = \frac{ \tan x }{ \cot x } * \tan^{2} x = \tan^{2} x * \tan^{2} x = \tan^{4} x$

10. VeroZarate

where did you get the second tan^2x

11. binarymimic

I divided tan^3 x into tan^x * tan^2 x

12. binarymimic

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13. VeroZarate

oh ok

14. binarymimic

So this is why we have the following equation so far: $\sec^{4} x = 1 + 2 \tan^{2} x + \tan^{4} x$

15. binarymimic

next step is to subtract both sides by tan^4 x

16. binarymimic

$\sec^{4} x - \tan^{4} x = 1 + 2 \tan^{2} x$

17. VeroZarate

and divide both side by2?

18. binarymimic

no there is a trick here now. notice the left side is the difference of two squares

19. binarymimic

$\sec^{4} x - \tan^{4} x = (\sec^{2} x)^{2} - (\tan^{2} x)^{2}$

20. binarymimic

remember that a^2 - b^2 = (a - b)(a + b) so we can now write the left side as: $\sec^{4} x - \tan^{4} x = (\sec^{2} x - \tan^{2} x)(\sec^{2} x + \tan^{2} x)$

21. VeroZarate

right you foil it?

22. binarymimic

we kind of did the reverse of FOIL, we factored it

23. binarymimic

So now we have: $(\sec^{2} x - \tan^{2} x)(\sec^{2} x + \tan^{2} x) = 1 + 2 \tan^{2} x$

24. binarymimic

but do you remember the identity sec^2 x = tan^2 x + 1

25. VeroZarate

yes but there is the 2?

26. binarymimic

we'll get to that :) so if sec^2 x = tan^2 x + 1 the if we subtract tan^2 from both sides we get sec^2 x - tan^2 x = 1

27. binarymimic

Then we see how

28. binarymimic

$(\sec^{2} x - \tan^{2} x)(\sec^{2} x + \tan^{2} x) = 1 + 2 \tan^{2} x$ becomes $(1)(\sec^{2} x + \tan^{2} x) = 1 + 2 \tan^2{x}$

29. VeroZarate

isnt it sec^4

30. binarymimic

No we factored $\sec^{4} x - \tan^{4} x$ into $(\sec^{2} x - \tan^{2} x)(\sec^{2} x + \tan^{2} x)$ because it is a difference of sqaures

31. VeroZarate

ok so when you subtract tan^2x?

32. binarymimic

There is a trig identity: $\sec^{2} x = 1 + \tan^{2} x$ i am saying, if we rewrite it as: $\sec^{2} x - \tan^{2} x = 1$ then the left factor on our left hand side reduces to just 1: $(\sec^{2} x - \tan^{2} x)(\sec^{2} x + \tan^{2} x) = (1)(\sec^{2} x + \tan^{2} x)$

33. binarymimic

so we are now left with $\sec^2{x} + \tan^{2} x = 1 + 2 \tan^{2} x$

34. VeroZarate

ok you had lost me. now i get it sorry

35. binarymimic

its a complicated problem compared to the last one o_o

36. VeroZarate

yes it is. i still have to do 5 more assignments. lol

37. binarymimic

so now we have $\sec^2{x} + \tan^{2} x = 1 + 2 \tan^{2} x$ can you see the next step ?

38. VeroZarate

substitute the sec^2x with 1+ tan^2x?

39. binarymimic

we could, or we could save a step and just subtract tan^2 x from both sides

40. binarymimic

if we subtract tan^2 x from both sides we get $\sec^{2} x = 1 + \tan^{2} x$

41. binarymimic

which is an identity

42. VeroZarate

ok

43. VeroZarate

but how does this equal to cotxsec^4x?

44. binarymimic

the whole idea is to reduce the original equation into a trig identity

45. binarymimic

if you follow the steps then that's precisely what we did

46. VeroZarate

to verify that it equals on both sides?

47. binarymimic

if you manipulate an equation using algebra then whatever you end up with is equal to what you started with.

48. VeroZarate

oh ok thanks so much!

49. binarymimic

np