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gjhfdfg

Verify the identity. Justify each step. tan (theta) + cot (theta) = 1 / sin (theta) cos (theta)

  • one year ago
  • one year ago

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  1. AbhimanyuPudi
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    tan(theta) = sin(theta) / cos(theta) and cot(theta) = cos(theta) / sin(theta) Substitute these in the above equation and you will prove it right...Any doubt?

    • one year ago
  2. gjhfdfg
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    How would I substitute the 'sin'.?

    • one year ago
  3. AbhimanyuPudi
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    tanx + cotx = sinx/cosx + cosx/sinx = \[( \sin x ^{2} + \cos ^{2} ) / (\sin x)(\cos x) = 1/(\sin x)(\cos x)\]

    • one year ago
  4. gjhfdfg
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    I'm so confused

    • one year ago
  5. AbhimanyuPudi
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    we know, \[(\sin x) ^{2} + (\cos x) ^{2} = 1\]

    • one year ago
  6. zepdrix
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    |dw:1349378735588:dw|

    • one year ago
  7. zepdrix
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    |dw:1349379018546:dw| This could be your next step perhaps. :) Still confused?

    • one year ago
  8. zepdrix
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    Yah you need to remember a lot of identities in trig, spend a good amount of time committing them to memory. Cause you'll do a lot of switching them back and forth in your class! :O

    • one year ago
  9. gjhfdfg
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    I kinda get it.? Sorry math is my weakest subject. :/ Some of the steps confuse me a little bit, I'm going to look at them again

    • one year ago
  10. gjhfdfg
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    So since tan (theta) = sin (theta) / cos (theta) , and cot (theta) = cos (theta) / sin (theta) , sin (theta) / cos (theta) + cos (theta) / sin (theta) must equal to 1 / sin (theta) cos (theta), then you get the common denominator of 1 / sin (theta) cos (theta) . Is this basically what your saying.?

    • one year ago
  11. zepdrix
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    |dw:1349382130808:dw| Yes :) Justifying each step. The last step would look like this. Justifying it with "Since" before we actually make the substitution with the identity.

    • one year ago
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