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burhan101 Group Title

Trig identities

  • 2 years ago
  • 2 years ago

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  1. burhan101 Group Title
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    \[\huge \frac{ 1 }{\sin x+1 }- \frac{ 1 }{ sinx-1 } = \frac{ 2 }{ \cos^{2}x }\]

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

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

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

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

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

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

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

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

    • 2 years ago
  10. burhan101 Group Title
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    ohhh

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

    • 2 years ago
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