## anonymous 3 years ago Verify that cot x sec4x = cot x + 2 tan x + tan3x

1. anonymous

so i know cotx is cos/sin

2. anonymous

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

3. anonymous

this one is hard but try this

4. anonymous

first divide everything by cot x

5. anonymous

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

6. anonymous

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. anonymous

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

8. anonymous

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

9. anonymous

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. anonymous

where did you get the second tan^2x

11. anonymous

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

12. anonymous

|dw:1357511455432:dw|

13. anonymous

oh ok

14. anonymous

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

15. anonymous

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

16. anonymous

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

17. anonymous

and divide both side by2?

18. anonymous

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

19. anonymous

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

20. anonymous

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. anonymous

right you foil it?

22. anonymous

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

23. anonymous

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

24. anonymous

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

25. anonymous

yes but there is the 2?

26. anonymous

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. anonymous

Then we see how

28. anonymous

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

29. anonymous

isnt it sec^4

30. anonymous

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. anonymous

ok so when you subtract tan^2x?

32. anonymous

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. anonymous

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

34. anonymous

ok you had lost me. now i get it sorry

35. anonymous

its a complicated problem compared to the last one o_o

36. anonymous

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

37. anonymous

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

38. anonymous

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

39. anonymous

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

40. anonymous

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

41. anonymous

which is an identity

42. anonymous

ok

43. anonymous

but how does this equal to cotxsec^4x?

44. anonymous

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

45. anonymous

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

46. anonymous

to verify that it equals on both sides?

47. anonymous

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

48. anonymous

oh ok thanks so much!

49. anonymous

np