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    #1

    Feb 6, 2011, 02:15 PM
    Verify identity
    tan(a-b) = (1-cot(a)tan(b)/cot(a)+tan(b))
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    #2

    Feb 6, 2011, 02:16 PM
    Verify identity
    (cotx)(secx)(sinx)=2-(tanx)(cosx)(cscx)
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    #3

    Feb 6, 2011, 05:08 PM
    sec(x)cos(x)=2-sin(x)csc(x) is that right
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    #4

    Feb 6, 2011, 05:57 PM
    (1)/(1+tan^(2)x)+(1)/(1+cot^(2)x)=1

    Replace 1+tan^(2)x with sec^(2)x using the identity sec^(2)(x)=1+tan^(2)(x).
    (1)/(sec^(2)x)+(1)/(1+cot^(2)x)

    Replace 1+cot^(2)x with csc^(2)x using the identity csc^(2)(x)=1+cot^(2)(x).
    (1)/(sec^(2)x)+(1)/(csc^(2)x)

    To add fractions, the denominators must be equal. The denominators can be made equal by finding the least common denominator (LCD). In this case, the LCD is secx^(2)cscx^(2). Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions.
    (1)/(sec^(2)x)*(csc^(2)x)/(csc^(2)x)+(1)/(csc^(2)x)*(sec^(2)x)/(sec^(2)x)

    Complete the multiplication to produce a denominator of secx^(2)cscx^(2) in each expression.
    (csc^(2)x)/(sec^(2)xcsc^(2)x)+(sec^(2)x)/(sec^(2)xcsc^(2)x)

    Combine the numerators of all expressions that have common denominators.
    (csc^(2)x+sec^(2)x)/(sec^(2)xcsc^(2)x)

    Reduce the expression ((cscx^(2)+secx^(2)))/(secx^(2)cscx^(2)) by removing a factor of from the numerator and denominator.
    (csc^(2)x+sec^(2)x)/(csc^(2)xsec^(2)x)

    Replace cscx with an equivalent expression (1)/(sin^(2)(x)) using the fundamental identities.
    (((1)/(sin^(2)(x))+sec^(2)x))/(csc^(2)xsec^(2)x)

    Replace secx with an equivalent expression (1)/(cos^(2)(x)) using the fundamental identities.
    (((1)/(sin^(2)(x))+(1)/(cos^(2)(x))))/(csc^(2)xsec^(2)x)

    To add fractions, the denominators must be equal. The denominators can be made equal by finding the least common denominator (LCD). In this case, the LCD is sin(x)^(2)cos(x)^(2). Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions.
    ((((1)/(sin^(2)(x))*(cos^(2)(x))/(cos^(2)(x))+(1)/(cos^(2)(x))*(sin^(2)(x))/(sin^(2)(x)))))/(csc^(2)xsec^(2)x)

    Complete the multiplication to produce a denominator of sin(x)^(2)cos(x)^(2) in each expression.
    ((((cos^(2)(x))/(sin^(2)(x)cos^(2)(x))+(sin^(2)(x))/(sin^(2)(x)cos^(2)(x)))))/(csc^(2)xsec^(2)x)

    Combine the numerators of all expressions that have common denominators.
    ((((cos^(2)(x)+sin^(2)(x))/(sin^(2)(x)cos^(2)(x)))))/(csc^(2)xsec^(2)x)

    Replace sin^(2)(x)+cos^(2)(x) with 1 using the identity sin^(2)(x)+cos^(2)(x)=1.
    ((((1)/(sin^(2)(x)cos^(2)(x)))))/(csc^(2)xsec^(2)x)

    Remove the parentheses from the numerator.
    (((1)/(sin^(2)(x)cos^(2)(x))))/(csc^(2)xsec^(2)x)

    Remove the parentheses around the expression (1)/(sin(x)^(2)cos(x)^(2)).
    ((1)/(sin^(2)(x)cos^(2)(x)))/(csc^(2)xsec^(2)x)

    Convert the csc^(2)x to its (sin)/(cos) equivalent.
    ((1)/(sin^(2)(x)cos^(2)(x)))/((((1)/(sin^(2)(x)))))

    Convert the sec^(2)x to its (sin)/(cos) equivalent.
    ((1)/(sin^(2)(x)cos^(2)(x)))/((((1)/(sin^(2)(x)))((1)/(cos^(2)(x)))))

    Replace sin(x) with an equivalent expression in the numerator.
    ((1)/(sin^(2)(x)cos^(2)(x)))/(((csc^(2)(x))((1)/(cos^(2)(x)))))

    Replace cos(x) with an equivalent expression in the numerator.
    ((1)/(sin^(2)(x)cos^(2)(x)))/((csc^(2)(x))(sec^(2)(x)))

    Remove the extra parentheses around the factors.
    ((1)/(sin^(2)(x)cos^(2)(x)))/(csc^(2)(x)(sec^(2)(x)))

    Multiply csc(x)^(2) by each term inside the parentheses.
    ((1)/(sin^(2)(x)cos^(2)(x)))/(csc^(2)(x)sec^(2)(x))

    Replace the expressions with an equivalent expression using the fundamental identities.
    ((1)/(sin^(2)(x)cos^(2)(x)))/(((1)/(sin^(2)(x)cos^(2)(x))))

    To divide by (1)/(sin(x)^(2)cos(x)^(2)), multiply by the reciprocal of the fraction.
    sin^(2)(x)cos^(2)(x)*(1)/(sin^(2)(x)cos^(2)(x))

    Remove the common factor of sin(x)^(2)cos(x)^(2) from the first term sin(x)^(2)cos(x)^(2) and the denominator of the second term (1)/(sin(x)^(2)cos(x)^(2)).
    1*1

    Multiply 1 by 1 to get 1.
    1
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    #5

    Feb 6, 2011, 05:58 PM
    Comment on idllotsaroms's post
    Is that right
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    #6

    Feb 6, 2011, 05:58 PM
    Comment on idllotsaroms's post
    NO!
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    #7

    Feb 6, 2011, 05:58 PM
    Comment on idllotsaroms's post
    Possibly
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    #8

    Feb 6, 2011, 06:03 PM
    Comment on idllotsaroms's post
    I'm not sure
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    #9

    Feb 6, 2011, 06:03 PM
    Comment on idllotsaroms's post
    Well everyone on this site is useless so this is the best u get
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    #10

    Feb 6, 2011, 06:03 PM
    Comment on idllotsaroms's post
    Okay

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