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eroshea Group Title

x=(tanA+cotA)^2 sinA-tan^2A

  • 2 years ago
  • 2 years ago

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  1. henpen Group Title
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    \[x=(tanA+cotA)^2 *(sinA-\tan^2A)\]This?

    • 2 years ago
  2. uzumakhi Group Title
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    is @henpen right?

    • 2 years ago
  3. eroshea Group Title
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    yes

    • 2 years ago
  4. henpen Group Title
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    \[(tanA+cotA)=(\frac{sinA}{cosA}+\frac{cosA}{sinA})=(\frac{sinAsinA}{cosAsinA}+\frac{cosAcosA}{sinAcosA})=\frac{1}{cosAsinA}\]

    • 2 years ago
  5. henpen Group Title
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    \[\because \cos^2(x)+\sin^2(x)=1\]

    • 2 years ago
  6. henpen Group Title
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    \[sinA-\tan^2A=\frac{sinAcos^2A}{\cos^2A}-\frac{\sin^2A}{\cos^2A}=\frac{sinAcos^2A-\sin^2A}{\cos^2A}\]

    • 2 years ago
  7. eroshea Group Title
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    give me a minute to analyze what you had illustrated

    • 2 years ago
  8. henpen Group Title
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    Multiplying,\[(\frac{1}{cosAsinA})^2*(\frac{sinAcos^2A-\sin^2A}{\cos^2A})=\frac{sinAcos^2A-\sin^2A}{\cos^4Asin^2A}\]

    • 2 years ago
  9. henpen Group Title
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    Which you can simplify if you want.

    • 2 years ago
  10. Zekarias Group Title
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    you must get this csc(x)-sec^2(x)

    • 2 years ago
  11. eroshea Group Title
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    owh.. i'm sorry .. you should find the value of x? the possible answers are: a. 4 b.3 c.2 d.1

    • 2 years ago
  12. henpen Group Title
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    \[cscAsec^2A-\sec^4A\]I must be mistaken.

    • 2 years ago
  13. eroshea Group Title
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    how will i arrive at the possible answers?

    • 2 years ago
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