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wampominater

  • one year ago

verify the trigonometric equation by substituting identities to match the right hand side with the left hand side: (sinx / 1-cosx) + (sinx / 1+cosx) = 2cscx

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  1. wampominater
    • one year ago
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    looks like this \[\frac{ sinx }{ 1-cosx } + \frac{ sinx }{ 1+cosx }=2cscx\] been stuck on this forever...

  2. maheshmeghwal9
    • one year ago
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    on left hand side take L.C.M.

  3. butterflydreamer
    • one year ago
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    start with your left hand side :) hint: You'll want put the two fractions over the same denominator .

  4. wampominater
    • one year ago
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    so, would i get them both under 1-cos^2x?

  5. wampominater
    • one year ago
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    over*

  6. butterflydreamer
    • one year ago
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    correct:)

  7. maheshmeghwal9
    • one year ago
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    It Becomes \[\frac{ sinx(1+cosx)+sinx(1-cosx) }{ \sin^2x }\]

  8. wampominater
    • one year ago
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    alright then do I distribute sin?

  9. butterflydreamer
    • one year ago
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    yyeeepp

  10. butterflydreamer
    • one year ago
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    distribute sinx :)

  11. wampominater
    • one year ago
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    \[\frac{ 2sinx }{ \sin^2x }\] like this?

  12. maheshmeghwal9
    • one year ago
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    yes

  13. butterflydreamer
    • one year ago
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    then from there, you can simplify the fraction and you'll get the answer (RHS) :)

  14. wampominater
    • one year ago
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    Ok so how do I simplify this, Sorry! this is where i have been stuck at.

  15. butterflydreamer
    • one year ago
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    oh you can divide the numerator and denominator by sinx

  16. wampominater
    • one year ago
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    so is it 2/sinx? And then just change it to csc?

  17. butterflydreamer
    • one year ago
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    likeee: |dw:1436251446732:dw|

  18. butterflydreamer
    • one year ago
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    and yes that's correct :)

  19. wampominater
    • one year ago
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    ahh ok THANK YOUUU!!!!!!!!!!

  20. butterflydreamer
    • one year ago
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    no problem :)!!

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