## anonymous one year ago Verify the trigonometric equation cot(x)sec^4(x)=cot(x)+2tan(x)+tan^3(x)?

1. anonymous

@Hero , @ganeshie8 @iambatman help please

2. SolomonZelman

are you sure it is not tan^2x ?

3. anonymous

1 second please let me make sure

4. SolomonZelman

oh, no nvm, it should be like that... but still make sure.

5. anonymous

it is how I initially wrote it

6. SolomonZelman

I would start from multiplying times tan(x) on both sides.

7. SolomonZelman

$$\large\color{black}{ \displaystyle \cot x\sec^4x=\cot x+2\tan x+\tan^3x }$$ $$\large\color{black}{ \displaystyle \sec^4x=1+2\tan^2x+\tan^4x }$$

8. anonymous

I have to focus on one side though, I am trying to prove that they are the same, not solve.

9. SolomonZelman

oh, okay, it would have been wonderful, but I guess not

10. SolomonZelman

$$\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\tan x+\tan^3x }$$ $$\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2x+\tan^4x\right) }$$ would you agree with this?

11. SolomonZelman

I factored out of cot(x).

12. anonymous

I have seen that everywhere I have looked, but I dont understand where the tan^2 or the Tan^4 come from

13. SolomonZelman

well, do you agree that: $$\cot(x) \times \tan(x)=1$$ $$\cot(x) \times \tan^2(x)=\tan(x)$$ $$\cot(x) \times \tan^3(x)=\tan^2(x)$$ $$\cot(x) \times \tan^4(x)=\tan^3(x)$$ and so on... $$\cot(x) \times \tan^{N}(x)=\tan^{N-1}(x)$$

14. SolomonZelman

if you don't understand why these are true, then ask

15. anonymous

I didn't know those identities. I knew that sin^2x + cos^2x =1 ....

16. SolomonZelman

well, do you understand why they are logically true?

17. dinamix

@SolomonZelman i want ask u how did u get sec^4(x) = 1 +2tan^2(x) +tan^4(x)

18. anonymous

Not really.

19. SolomonZelman

I didn't get that yet, but we will get that.

20. SolomonZelman

$$\large\color{black}{ \displaystyle (x+1)^2=x^2+2x+1 }$$ right?

21. dinamix

we have only sec(x) = 1/cos(x)

22. SolomonZelman

turmoilx2, do you agree that: $$\cot(x) \times \tan(x)=1$$

23. anonymous

I suppose... if you tell me that it's true

24. anonymous

@dinamix . Are you in FLVS?

25. SolomonZelman

$$\cot(x) \times \tan(x)=1$$ $$\dfrac{\cos(x)}{\sin(x)}\times \dfrac{\sin(x)}{\cos(x)}=1$$

26. anonymous

I understand that.

27. SolomonZelman

28. anonymous

Yes!

29. SolomonZelman

ok, my comp glitched:)

30. SolomonZelman

$$\cot(x) \times \tan(x)=1$$ So, now, we will be multiplying times tan(x) on both sides: $$\cot(x) \times \tan^2(x)=\tan(x)$$ $$\cot(x) \times \tan^3(x)=\tan^2(x)$$ $$\cot(x) \times \tan^4(x)=\tan^3(x)$$ and so on... $$\cot(x) \times \tan^{N}(x)=\tan^{N-1}(x)$$ -------------------------------- Is this better now, why the list of these identites is true?

31. dinamix

@turmoilx2 no mate

32. SolomonZelman

turmoilx, when you read, reply me if you got what I elaborates so far in my last reply.

33. anonymous

I understand why the identities are true now

34. SolomonZelman

Ok, so now back to our question;)

35. SolomonZelman

$$\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\tan x+\tan^3x }$$ Now, using these identies, I am factoring the left side out of $$\cot(x)$$. $$\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2x+\tan^4x\right) }$$ Do you get how I am factoring?

36. anonymous

37. SolomonZelman

let me do an intermediate step in blue: $$\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\tan x+\tan^3x }$$ $$\large\color{blue}{ \displaystyle \cot x \sec^4x=\cot x\cdot 1+2\cot x\tan^2x+\cot x\tan^4x }$$ $$\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2x+\tan^4x\right) }$$ is it better now?

38. SolomonZelman

after the blue, I factor out of $$\cot(x)$$. If you still have questions about how I factored, then please ask.

39. anonymous

Im still lost. Sorry if I'm being slow here, online math classes are not for me.

40. SolomonZelman

(it is fine if you don't I can attempt to explain a little better than what I did. I am not a very good tutor after all).

41. anonymous

I dont understand how you can just multiply the tan by cot and I dont know how to get the exponents for the tangents still. I understand the identities you mentioned above but I dont know how they apply to this first step if that makes sense...

42. anonymous

oh wait...

43. anonymous

Is it allowed because Cotxtanx is the same as multiplying by 1 which wouldnt change the equation?

44. SolomonZelman

yes

45. anonymous

I think I understand some more now...

46. SolomonZelman

but I would still just in case post my explanation on that as throuhg as I can. I will try not to take too much time.

47. anonymous

Yes. That would be appreciated, thank you!

48. SolomonZelman

$$\large\color{black}{ \displaystyle \cot x \sec^4x=\color{blue}{\cot x}+\color{green}{2\tan x}+\color{red}{\tan^3x} }$$ -----------------------------------------------$${\$0.8em]}$$ $$\large\color{blue}{ \displaystyle \cot x =\cot x }$$ (We know that)$${\\[0.8em]}$$ -----------------------------------------------$${\\[0.8em]}$$ $$\large\color{green}{ \displaystyle \tan x =\cot x\tan^2x }$$$${\\[0.8em]}$$ (As we showed before in the list of those identites)$${\\[0.8em]}$$ $$\large\color{green}{ \displaystyle \tan x =\cot x\tan^2x }$$$${\\[0.8em]}$$ And then multiply times 2 on both sides$${\\[0.8em]}$$ $$\large\color{green}{ \displaystyle 2\tan x =2\cot x\tan^2x }$$ $${\\[0.8em]}$$ -----------------------------------------------$${\\[0.8em]}$$ $$\large\color{red}{ \displaystyle \tan^3x =\cot x\tan^4x }$$$${\\[0.8em]}$$ (Also, as we showed before in the list of those identites)$${\\[0.8em]}$$ ----------------------------------------------- So it comes out that: $$\large\color{black}{ \displaystyle \cot x \sec^4x=\color{blue}{\cot x}+\color{green}{2\cot x\tan^2 x}+\color{red}{\cot x\tan^4x} }$$ 49. SolomonZelman take your time. 50. SolomonZelman When you are done reading, tell me if you understand how I got the following equation below: $$\large\color{black}{ \displaystyle \cot x \sec^4x=\color{blue}{\cot x}+\color{green}{2\cot x\tan^2 x}+\color{red}{\cot x\tan^4x} }$$ 51. anonymous Here is what I have so far: \[CotxSec^4x=Cotx+2Tanx+Tan^3x$ $CotxSec^4x=Cotx+2(CotxTan^2x)+(CotxTan^4x)$ $CotxSec^4x=Cotx+Cot(2Tan^2x+tan^4x) Correct??? 52. anonymous oops \[CotxSec^4x=Cotx+Cot(2Tan^2x+tan^4x)$

53. SolomonZelman

$$\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\cot x\tan^2 x+\cot x\tan^4x}$$ Rather, take out $$\cot(x)$$ from the entire right side. $$\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2 x+\tan^4x\right)}$$

54. SolomonZelman

did you get everything I have done so far/

55. anonymous

$cotxsec4x=cotx((((1))))+2\tan2x+\tan4x)$ So Im assuming this "1" is from removing the cotagents... but how?

56. SolomonZelman

well when I factor cot(x) out of cot(x), then I get 1.

57. SolomonZelman

$$\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\cot x\tan^2 x+\cot x\tan^4x}$$ I am factoring out of $$\cot(x)$$. And I factor every term on the left side. $$\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2 x+\tan^4x\right)}$$ still questions about that factoring out?

58. anonymous

I think what I dont get is why Cotx+2CotxTan^2x doesnt become 1+2Tanx instead of 1+ 2Tan^2x

59. SolomonZelman

2tan^2x is what I got, didn't I?

60. SolomonZelman

oh, because you have 2cot(x)tan²(x), and thus when you take out the cot(x) part, you remain with 2tan²(x).

61. anonymous

I'll just go with it. Cot can be factored from itself to equal 1.

62. SolomonZelman

yes, true

63. SolomonZelman

$$\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2 x+\tan^4x\right)}$$ then, you know that: $$\color{red}{({\rm w}+1)^2={\rm w}^2+2{\rm w}+1}$$ So, then look at the parenthesis on the right side and tell me how THAT can be factors.

64. SolomonZelman

$$\large ($$$$\color{red}{\rm w}$$ in our case is $$\tan^2x$$$$\large )$$

65. anonymous

$(w+1)^2=w^2+1^2$ I know that this much is true.

66. SolomonZelman

no, that is not true

67. SolomonZelman

w²+1 would be w²+1², BUT (w+1)² is different:

68. SolomonZelman

$$(w+1)^2=(w+1)(w+1)=w(w+1)+1(w+1) \\ =w^2+w+w+1=w^2+2w+1$$

69. SolomonZelman

so, therefore we know that: $$(w+1)^2=w^2+2w+1$$

70. SolomonZelman

ok, tell me if you get my very last post...

71. anonymous

I get it now. Mind blown

72. anonymous

You are very patient. Thanks

73. SolomonZelman

Ok, so now if you look at, and understand the fact that: $$(w+1)^2=w^2+2w+1$$ Then you should also know that: $$(\tan^2x+1)^2=\left(\tan^2x\right)^2+2\left(\tan^2x\right)+1$$ - - - - - - - - - - - - - - - - - - - - - - - - And then take a look at your problem: $$\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2 x+\tan^4x\right)}$$ and it should follow that: $$\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+\tan^2 x\right)^2}$$

74. SolomonZelman

By the way, it is not a problem.

75. anonymous

and then (1+tan^2x)^2 = (sec^2x)^2 which then equals sec^4x

76. SolomonZelman

Yes, that is correct!

77. anonymous

THANK YOU SO MUCH!

78. SolomonZelman

And then, we have verified the original equation to be an identity.

79. SolomonZelman

You are welcome:)