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kaylala

  • 11 months ago

topic: PROVING IDENTITIES - help me please (SEE COMMENTS) (trigonometry)

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  1. kaylala
    • 11 months ago
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    @Loser66 help

  2. SolomonZelman
    • 11 months ago
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    post your question please.

  3. SolomonZelman
    • 11 months ago
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    ( Loser66 is offline )

  4. kaylala
    • 11 months ago
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    (1)/(sec x - tan x) = sec x + tan x

  5. kaylala
    • 11 months ago
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    i noticed

  6. SolomonZelman
    • 11 months ago
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    \[\huge\color{blue}{ \frac{1}{\sec(x)-\tan(x)} =\sec(x)+\tan(x) } \]

  7. SolomonZelman
    • 11 months ago
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    \[\huge\color{blue}{ 1 =(\sec(x)+\tan(x) )(\sec(x)-\tan(x))} \]

  8. SolomonZelman
    • 11 months ago
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    I multiplied both sides times sec(40)-tan(40)

  9. kaylala
    • 11 months ago
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    why? is that possible? i thought we can only manipulate 1 side?

  10. kaylala
    • 11 months ago
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    shouldn't the other side be stable?

  11. SolomonZelman
    • 11 months ago
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    Oh, then I take what I wrote back.

  12. kaylala
    • 11 months ago
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    okay

  13. SolomonZelman
    • 11 months ago
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    \[\huge\color{blue}{ \frac{1}{\sec(x)-\tan(x)}=\sec(x)+\tan(x) } \] LEFT SIDE:\ top and bottom times sec(x)+tan(x) \[\huge\color{blue}{ \frac{\sec(x)-\tan(x)}{\sec^2(x)-\tan^2(x)}=\sec(x)+\tan(x) } \] Now, use the identity for the bottom, \[\sec^2x=\tan^2x+1\]

  14. SolomonZelman
    • 11 months ago
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    Oh, the top should say + not - sorry.

  15. SolomonZelman
    • 11 months ago
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    \[\huge\color{blue}{ \frac{\sec(x) \huge\color{red}{ + } \tan(x)}{\sec^2(x)-\tan^2(x)}=\sec(x)+\tan(x) } \]

  16. SolomonZelman
    • 11 months ago
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    HUGE HINT: \[\sec^2x=\tan^2x+1~~~~~~~ther efore~~~~~\sec^2x-\tan^2x=1\]

  17. kaylala
    • 11 months ago
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    i do not get it. sorry

  18. SolomonZelman
    • 11 months ago
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    Do you get what I did before. the last huge post in blue with a red plus ?

  19. kaylala
    • 11 months ago
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    nope but hey i found this: http://symbolab.com/solver/trigonometric-identity-calculator/prove%20%5Cfrac%7B1%7D%7B%5Csec(x)-%5Ctan(x)%7D%3D%5Csec(x)%2B%5Ctan(x) but the answer doesnt seem to be equal is there a way you could make it equal???

  20. SolomonZelman
    • 11 months ago
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    Don't try to look it up else where, follow me. lets start over, ok? \[\huge\color{green} { \frac{1}{\sec(x)-\tan(x)}=\sec(x)+\tan(x) }\] I am working the left side: 1) Multiply top and bottom times sec(x)+tan(x)

  21. kaylala
    • 11 months ago
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    ok

  22. SolomonZelman
    • 11 months ago
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    \[\color{green} { \frac{\sec(x)+\tan(x)}{(~~\sec(x)-\tan(x)~~)~(~~\sec(x)+\tan(x)~~)}=\sec(x)+\tan(x) }\] \[\color{green} { \frac{\sec(x)+\tan(x)}{\sec^2(x)-\tan^2(x)}=\sec(x)+\tan(x) }\] tell me if you don't get it.

  23. kaylala
    • 11 months ago
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    okay

  24. SolomonZelman
    • 11 months ago
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    Now lets use an identity. \[\tan^2(x)+1=\sec^2(x)\] let's draw some conclusions. \[\tan^2(x)+1\color{blue} { -\tan^2(x) } =\sec^2x\color{blue} { -\tan^2(x) }\]\[1=\sec^2x-\tan^2x\] look at the denominator, what is it now equal to? has your identity been verified?

  25. kaylala
    • 11 months ago
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    yes. got it thanks @SolomonZelman

  26. SolomonZelman
    • 11 months ago
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    You welcome, no problem, just knowing identities that's all the skill involved.

  27. kaylala
    • 11 months ago
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    and wow. that was shorter

  28. SolomonZelman
    • 11 months ago
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    Yeah, I explained the same thing, but in a shorter way.

  29. kaylala
    • 11 months ago
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    what's the technique?

  30. kaylala
    • 11 months ago
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    how come you're so good at this?

  31. SolomonZelman
    • 11 months ago
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    what do you mean, the technique for what? Ohhh, I took trig last year, and got 97. I am just good at math and bad at other staff.

  32. kaylala
    • 11 months ago
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    technique in answering / proving these trigonometric identities? you did it so fast and in a shorter and much efficient way that's really cool WOW! Congratz! good for you

  33. SolomonZelman
    • 11 months ago
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    Thank you! I am sure that if you try to do a lot of practice problems you will master them.

  34. kaylala
    • 11 months ago
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    i really hope so. failing trigonometry is the last thing that i'd want

  35. SolomonZelman
    • 11 months ago
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    WHAT?! FAILING?! what is your current grade may I ask ?

  36. kaylala
    • 11 months ago
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    last semester i got 2.0 in algebra now i'm taking trigonometry and it do is very hard. i dont have my grade yet since the 2nd semester just started.

  37. kaylala
    • 11 months ago
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    1 being the highest 3 - pass 5 - fail

  38. kaylala
    • 11 months ago
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    @SolomonZelman ^

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