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feldy90

  • 2 years ago

Is the cotan(theta) function the same as tan^-1(theta)?

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  1. uzumakhi
    • 2 years ago
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    \[\cot \theta = \frac{ 1 }{ \tan \theta }\]

  2. feldy90
    • 2 years ago
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    So that's a yes? If I'm entering it into my calculator (there is no cot function), I enter\[\cot(\theta) = \tan^{-1}(\theta)\]

  3. uzumakhi
    • 2 years ago
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    you can't write it as \[\tan ^{-1} \theta \]

  4. feldy90
    • 2 years ago
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    Isn't that the same thing...? Doesn't: \[\frac{ 1 }{ \tan(\theta) } = \tan ^{-1}(\theta)\]

  5. uzumakhi
    • 2 years ago
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    no my friend

  6. feldy90
    • 2 years ago
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    :( okay. Thanks

  7. irene22988
    • 2 years ago
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    nope. to find cot(theta), do this: 1/(tan(theta))

  8. uzumakhi
    • 2 years ago
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    when we take tan function from one side of equality to other then we write tan^-1

  9. feldy90
    • 2 years ago
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    So you use tan^-1 in something like... To solve for x if tan(x)=a, then x=tan^-1(a)

  10. uzumakhi
    • 2 years ago
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    yes you are right

  11. feldy90
    • 2 years ago
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    Thank you :)

  12. uzumakhi
    • 2 years ago
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    welcome

  13. feldy90
    • 2 years ago
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    Can you say\[\csc ^{4}(\theta) = \frac{ 1 }{ \sin ^{4}(\theta) }\]

  14. irene22988
    • 2 years ago
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    yep. in calculator, that would be: 1/(sin(theta))^4

  15. shubhamsrg
    • 2 years ago
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    -1 is a special case in trigonometric functions.. -1 corresponds to inverse function.. thus for 1/sin, we use cosec and likewise..

  16. feldy90
    • 2 years ago
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    Thanks I get it now :)

  17. shubhamsrg
    • 2 years ago
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    hmm,,glad you do!

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