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sabika13

  • 2 years ago

How do I prove this by using trig identities: (2sin^2x-1)/(sinxcosx)/=tanx-cotx

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  1. ghazi
    • 2 years ago
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    \[2 \sin^2 x-1= 2\sin^2x-(\sin^2x+\cos^2x)= \sin^2x-\cos^2x \]

  2. sabika13
    • 2 years ago
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    how does that equal to sin^2x-cos^2x

  3. ghazi
    • 2 years ago
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    sin^2 x + cos^2x=1

  4. sabika13
    • 2 years ago
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    sin^2x-cos^2x/sin^2xcosx..now what do i do

  5. ghazi
    • 2 years ago
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    \[\frac{ \sin^2 }{ \sin x *\cos x }- \frac{ \cos^2 x }{ sinx * \cos x }= tanx - \cot x\] clear?

  6. ghazi
    • 2 years ago
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    @sabika13 hope that helps you

  7. sabika13
    • 2 years ago
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    how doss sin^2/sinxcosx=tanx?

  8. sabika13
    • 2 years ago
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    does

  9. ghazi
    • 2 years ago
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    |dw:1352223957740:dw| how about now?

  10. ghazi
    • 2 years ago
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    still confused?

  11. ghazi
    • 2 years ago
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    |dw:1352224079572:dw|

  12. sabika13
    • 2 years ago
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    ohhhhh thank you so much, i wrote my denominator wrong, thats why i was confused

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