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

1. VeroZarate Group Title

so i know cotx is cos/sin

2. VeroZarate Group Title

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

3. binarymimic Group Title

this one is hard but try this

4. binarymimic Group Title

first divide everything by cot x

5. binarymimic Group Title

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

6. binarymimic Group Title

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 Group Title

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

8. binarymimic Group Title

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

9. binarymimic Group Title

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 Group Title

where did you get the second tan^2x

11. binarymimic Group Title

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

12. binarymimic Group Title

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13. VeroZarate Group Title

oh ok

14. binarymimic Group Title

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

15. binarymimic Group Title

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

16. binarymimic Group Title

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

17. VeroZarate Group Title

and divide both side by2?

18. binarymimic Group Title

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

19. binarymimic Group Title

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

20. binarymimic Group Title

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 Group Title

right you foil it?

22. binarymimic Group Title

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

23. binarymimic Group Title

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

24. binarymimic Group Title

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

25. VeroZarate Group Title

yes but there is the 2?

26. binarymimic Group Title

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 Group Title

Then we see how

28. binarymimic Group Title

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

29. VeroZarate Group Title

isnt it sec^4

30. binarymimic Group Title

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 Group Title

ok so when you subtract tan^2x?

32. binarymimic Group Title

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 Group Title

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

34. VeroZarate Group Title

ok you had lost me. now i get it sorry

35. binarymimic Group Title

its a complicated problem compared to the last one o_o

36. VeroZarate Group Title

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

37. binarymimic Group Title

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

38. VeroZarate Group Title

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

39. binarymimic Group Title

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

40. binarymimic Group Title

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

41. binarymimic Group Title

which is an identity

42. VeroZarate Group Title

ok

43. VeroZarate Group Title

but how does this equal to cotxsec^4x?

44. binarymimic Group Title

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

45. binarymimic Group Title

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

46. VeroZarate Group Title

to verify that it equals on both sides?

47. binarymimic Group Title

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

48. VeroZarate Group Title

oh ok thanks so much!

49. binarymimic Group Title

np