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anonymous

  • one year ago

Verify the identity. ((cos x)/1+sin x) + ((1+sinx)/cos x) =2secx

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  1. anonymous
    • one year ago
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    \[\frac{ cosx }{ 1+\sin x }+\frac{ 1+\sin x }{ \cos x }=2secx\]

  2. anonymous
    • one year ago
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    I don't know if I'm doing this right. I used LCD.

  3. anonymous
    • one year ago
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    \[LEFT SIDE: \frac{ \cos^2x +(1+\sin x)(1+\sin x)}{ \cos(1+sinx) }\]

  4. anonymous
    • one year ago
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    \[ \frac{ \cos^2x +1+2sinx+sin^2x}{ \cos(1+sinx) }\]

  5. Mertsj
    • one year ago
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    \[\frac{(\cos x)(1-\sin x)}{(1+\sin x)(1-\sin x)}+\frac{(1+\sin x)(\cos x)}{\cos ^2x}=\frac{\cos x-\cos x \sin x}{1-\sin ^2x}+\frac{\cos x+\cos x \sin x}{\cos ^2x}\]

  6. Nnesha
    • one year ago
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    change cos^2 by 1-sin^2x identity

  7. anonymous
    • one year ago
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    Wait, am I doing it right?

  8. Mertsj
    • one year ago
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    \[\frac{2\cos x}{\cos ^2x}=\frac{2}{\cos x}=2\sec x\]

  9. anonymous
    • one year ago
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    @Mertsj What did you do first?

  10. Mertsj
    • one year ago
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    Multiply first fraction by (1-sinx)/(1-sinx)

  11. Mertsj
    • one year ago
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    Multiply second fraction by cosx/cosx

  12. anonymous
    • one year ago
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    Ohhh is it like multiplying the conjugate?

  13. Mertsj
    • one year ago
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    it's really setting up cos^2x in both denominators

  14. Mertsj
    • one year ago
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    Because 1-sin^2x=cos^2x

  15. Nnesha
    • one year ago
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    \[\huge\rm \frac{ \color{Red}{\cos^2x} +1 +2sinx + \sin^2 }{ cosx(1+sinx) }\] \[\large\rm \frac{ \color{Red}{1-sin^2x} +1 +2sinx + \sin^2 }{ cosx(1+sinx) }\] sin^2 cancel each other out you will get \[\large\rm \frac{ \color{Red}{1\cancel{-sin^2x}} +1 +2sinx +\cancel{ \sin^2} }{ cosx(1+sinx) }\]

  16. anonymous
    • one year ago
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    Ohhh so what I was doing is wrong. Haha! Thanks!

  17. Nnesha
    • one year ago
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    i don't think so..

  18. Mertsj
    • one year ago
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    Not wrong, such not very effective.l

  19. anonymous
    • one year ago
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    Like it the solution would be longer? :D

  20. Nnesha
    • one year ago
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    \[\huge\rm \frac{ 2+2sinx }{ cosx(1+sinx) }\] take out 2 \[\rm \frac{ 2(1+sinx) }{ cosx(1+sinx)}\]

  21. anonymous
    • one year ago
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    There's no "it" in my last sentence. Haha.

  22. Mertsj
    • one year ago
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    Typically there are a variety of ways to solve these identities.

  23. anonymous
    • one year ago
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    Ohhhh @Nnesha thank you so much! :) You did the whole solution for me.

  24. Nnesha
    • one year ago
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    true but i think the way you were doing is easy :D

  25. anonymous
    • one year ago
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    That's why this is harder because there are so many ways to do it and it's so frustrating haha!

  26. Nnesha
    • one year ago
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    and like mertsj said there are more than 2 ways to verify the identities

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