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    help verifying identities

    are each of these an identity? show how...

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  2. #2
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    Quote Originally Posted by adam456006 View Post
    are each of these an identity? show how...

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    1. Use the difference of two squares:
    <br />
\frac{sin(u)(1+cos(u)}{(1-cos(u))(1+cos(u)} = \frac{(sin(u)(1+cos(u))}{1-cos^2(u)}

    1-cos^2(u) = sin^2(u)

    \frac{1+cos(u)}{sin(u)} so that's an identity
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    I was working with # 2 and I don't think it is an identity, but I'm not sure....
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    Hello, adam456006!

    Are each of these an identity?

    (1)\;\;\frac{\sin u}{1-\cos u} - \frac{1+\cos u}{\sin u } \:=\:0
    Get a common denominator and subtract . . .

    \frac{\sin u}{1-\cos u} \cdot{\color{blue}\frac{\sin u}{\sin u}} - \frac{1+\cos u}{\sin u}\cdot{\color{blue}\frac{1-\cos u}{1-\cos u}} \;= \;\frac{\sin^2\! u - \overbrace{(1-\cos^2\!u)}^{\text{This is }\sin^2\!u}}{\sin u(1-\cos u)}

    . . = \;\frac{\sin^2\!u-\sin^2\!u}{\sin u(1-\cos u)} \;=\;\frac{0}{\sin u(1-\cos u)} \;=\;0

    It is an identity.




    (2)\;\;\frac{\csc^2\!x - \cot^2\!x}{\sec x} \:=\:\sin x

    We have: . \frac{\overbrace{\csc^2\!x - \cot^2\!x}^{\text{This is 1}}}{\sec x} \;=\;\frac{1}{\sec x} \;=\;\cos x

    It is not an identity.




    (3)\;\;\frac{\cos x - \cos y}{\sin x + \sin y} + \frac{\sin x - \sin y}{\cos x + \cos y} \:=\:0
    Get a common denominator and add . . .

    \frac{\cos x - \cos y}{\sin x + \sin y}\cdot{\color{blue}\frac{\cos x + \cos y}{\cos x + \cos y}} + \frac{\sin x - \sin y}{\cos x + \cos y}\cdot{\color{blue}\frac{\sin x + \sin y}{\sin x + \sin y}}

    . . = \;\frac{\cos^2\!x - \cos^2\!y + \sin^2\!x - \sin^2\!y}{(\sin x+\sin y)(\cos x + \cos y)} \;=\;\frac{\overbrace{(\sin^2\!x +\cos^2\!x)}^{\text{This is 1}} - \overbrace{(\sin^2\!y + \cos^2\!y)}^{\text{This is 1}}}{(\sin x + \sin y)(\cos x + \cos y)}

    . . = \;\frac{1-1}{(\sin x+\sin y)(\cos x + \cos y)} \;=\;\frac{0}{(\sin x+\sin y)(\cos x+\cos y)} \;=\;0


    It is an identity.

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