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anonymous

  • 4 years ago

(1+SINX)^(1/2) 1+sinx -------------= ------- (1-sinx)^(1/2) |cosx|

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  1. anonymous
    • 4 years ago
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    verify

  2. anonymous
    • 4 years ago
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    okay, you have what exactly? this? \[\sqrt{\frac{1+sinx}{1-sin(x)}}=\frac{1+sin(x)}{|cos(x)|}\text{?}\]

  3. anonymous
    • 4 years ago
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    correct

  4. dumbcow
    • 4 years ago
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    multiply top and bottom by conjugate: sqrt(1+sin x) \[\frac{\sqrt{1+sinx}}{\sqrt{1-sinx}}*\frac{\sqrt{1+sinx}}{\sqrt{1+sinx}} = \frac{1+sinx}{\sqrt{1-\sin^{2} x}} = \frac{1+sinx}{cosx}\]

  5. campbell_st
    • 4 years ago
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    rationalise the denominator by multiplying by 1 + sin(x) will result in \[\sqrt{(1+\sin(x)^2/1 - \sin^2(x)}\] denominator is cos^2(x) while square root of the numerator is (1+sin(x)

  6. anonymous
    • 4 years ago
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    after the sqrt gets off the top wouldn't it go off the bottom as well

  7. campbell_st
    • 4 years ago
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    the reason its absolute value is that the square root will give a positive and negative answer.. hence the abs value

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