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burhan101

  • 2 years ago

Trig identities

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  1. burhan101
    • 2 years ago
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    \[\huge \frac{ 1 }{\sin x+1 }- \frac{ 1 }{ sinx-1 } = \frac{ 2 }{ \cos^{2}x }\]

  2. burhan101
    • 2 years ago
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    \[\huge \frac{ 1 }{ sinx+1 } - \frac{ 1 }{ sinx-1 } = \frac{ 2 }{ 1-\sin^2x }\]

  3. tkhunny
    • 2 years ago
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    You're going to have to add the ones on the left before you see it.

  4. burhan101
    • 2 years ago
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    what do you mean ? :S

  5. wio
    • 2 years ago
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    \[\Large \frac{a}{b} - \frac{c}{d} = \frac{ad-cb}{bd} \]

  6. burhan101
    • 2 years ago
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    for the right side i get this : \[\huge \frac{ (sinx-1)-(sinx+1) }{ 1-\sin^2x }\]

  7. irkiz
    • 2 years ago
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    dont make your life hard combine the left hand side into 1 fraction first

  8. irkiz
    • 2 years ago
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    then multiply the denominators to remove the fraction

  9. irkiz
    • 2 years ago
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    then it would be extremely easy to solve

  10. burhan101
    • 2 years ago
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    ohhh

  11. irkiz
    • 2 years ago
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    when u make the LHS into one fraction, u will get something like \[\frac{ a }{ b }=\frac{ c }{ d }\] \[ad = cb\]

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