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anonymous

  • one year ago

Verify the equation by substituting identities to match the right side to the left side. cot x sec4x = cot x + 2 tan x + tan3x

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  1. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \cot x \sec^4x = \cot x + 2 \tan x + \tan^3x }\) ?

  2. SolomonZelman
    • one year ago
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    Or, \(\large\color{black}{ \displaystyle \cot x ~\sec(4x) = \cot x + 2 \tan x + \tan(3x ) }\) ?

  3. anonymous
    • one year ago
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    The first one. :)

  4. SolomonZelman
    • one year ago
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    cool.

  5. anonymous
    • one year ago
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    So I'm trying to do the right side.

  6. anonymous
    • one year ago
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    And first I changed cot x to 1/tan x, and then cos x/sin x

  7. anonymous
    • one year ago
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    HI

  8. anonymous
    • one year ago
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    I'll try to write what I did first. :)

  9. anonymous
    • one year ago
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    RIGHT SIDE: \[\frac{ \cos x }{ \sin x }+\frac{ 2 \sin x }{ \cos x }+\frac{ \sin ^{3}x }{ \cos ^{3}x }\]

  10. anonymous
    • one year ago
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    And then I'm kinda stuck with it.

  11. anonymous
    • one year ago
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    I used the reciprocal and quotient identities by the way.

  12. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \cot x \sec^4x = \cot x + 2 \tan x + \tan^3x }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \cos x }{ \sin x }+\frac{ 2 \sin x }{ \cos x }+\frac{ \sin ^{3}x }{ \cos ^{3}x } }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \cos^4x }{ \sin x \cos^3x}+\frac{ 2 \cos^2x\sin^2x }{ \cos^3x\sin x }+\frac{ \sin ^{4}x }{ \cos ^{3}x \sin x} }\)

  13. anonymous
    • one year ago
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    I may sound dumb to you, but can you explain how you get that?

  14. SolomonZelman
    • one year ago
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    then combine the fractions and use (a+b)²=a²+2ab+b²

  15. SolomonZelman
    • one year ago
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    I just found the common denominator to each fraction.

  16. anonymous
    • one year ago
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    Ohhh. Thank you!

  17. anonymous
    • one year ago
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    I will work on it right now. So I will close this question. I'll just ask another question if I'm stuck with something. :)

  18. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \cos^4x }{ \sin x \cos^3x}+\frac{ 2 \cos^2x\sin^2x }{ \cos^3x\sin x }+\frac{ \sin ^{4}x }{ \cos ^{3}x \sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \cos^4x + 2 \cos^2x\sin^2x + \sin ^{4}x }{ \cos ^{3}x \sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \left(\cos^2x+\sin^2x\right)^2 }{ \cos ^{3}x \sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \left(1\right)^2 }{ \cos ^{3}x \sin x} }\)

  19. anonymous
    • one year ago
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    Please don't tell me the answer. :( I'm trying to know how to do it haha.

  20. SolomonZelman
    • one year ago
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    oh, sorry, go ahead.

  21. anonymous
    • one year ago
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    \[\cot x \sec ^{4}x=\frac{ 1 }{ \cos ^{3}xsin x }\] @SolomonZelman What's next?

  22. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x \sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x }\cdot\frac{1}{\sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{4}x }\cdot\frac{\cos x}{\sin x} }\)

  23. SolomonZelman
    • one year ago
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    then the rest is even easier...

  24. anonymous
    • one year ago
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    Why did it become like that?

  25. anonymous
    • one year ago
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    The third one I mean.

  26. SolomonZelman
    • one year ago
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    i multiplyied by cos(x)/cos(x)

  27. anonymous
    • one year ago
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    Why? Sorry if I keep on bugging you. Because what I did was I turned 1/sin x to csc x.

  28. SolomonZelman
    • one year ago
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    oh, don't .... and it's ok

  29. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x \sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x }\cdot\frac{1}{\sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x }\cdot\frac{1}{\sin x} \cdot \frac{\cos x}{\cos x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x\cdot \cos x }\cdot\frac{1\cdot \cos x }{\sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{4}x }\cdot\frac{\cos x}{\sin x} }\)

  30. anonymous
    • one year ago
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    And from the third one, it will become the equation on the left side, right?

  31. SolomonZelman
    • one year ago
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    yes, because: 1/cos\(^4\)x = ? cos(x)\(/\)sin(x) = ?

  32. anonymous
    • one year ago
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    sec^4x and cot x :) Thank you so much for guiding me! I can now do the other ones.

  33. SolomonZelman
    • one year ago
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    Yeah.... the main trick was in finding the common denominator and noticing a²+2ab+b²=(a+b)² you welcome, and I got a prayer to attend right now. cu

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