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anonymous

  • 5 years ago

An identity question.

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  1. anonymous
    • 5 years ago
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    (1-sin x)/(1+sin x) = (sec x - tan x)^2

  2. anonymous
    • 5 years ago
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    Working with the right side only. sec x = 1/cosx tan x = sinx/cosx Then you have a common denominator of cosx so you would have (1 - sinx) all over cos x then that is ^2. if you square both the top and the bottom you get (1 - sinx)(1 - sinx) all over cos^2x Do you know the identity sin^2x + cos^2x = 1 (This is the most important identity you should know) Solving this for cos^2x you get cos^2x = 1 - sin^2x... So then you substitute that in for cos^2x on the bottom and you get (1 - sinx)(1-sinx) over 1 - sin^2x But... the bottom is the difference of two squares so the bottom becomes (1 + sinx) (1 - sinx) ... Then you should see that the 1 - sinx cancels on the top and the bottom to leave the right side with 1 - sinx over 1 + sinx

  3. anonymous
    • 5 years ago
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    Well to add the first two fractions you need to get the LCD, for these two fractions it would be this: (6)7/(6)9 + (9)1/(9)6

  4. anonymous
    • 5 years ago
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    opps

  5. anonymous
    • 5 years ago
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    Thank you very much! It's a relief to know that I got half way through correctly!

  6. anonymous
    • 5 years ago
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    Really important to practice these. It is always easy to see when someone else does it... But.. It is important that you can do it by yourself.. Keep working at it.

  7. anonymous
    • 5 years ago
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    Thank you, I will. XD

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