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Math Help - Trig identity problem

  1. #1
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    Trig identity problem

    Prove -
    1)  sec (u) + tan (u) = \frac {1} {sec(u) - tan(u)}

    2)  sec (u) + cosec (u)cot(u) = sec(u)cosec^2(u)

    3)  sin^2 (u)(1+sec^2 (u) ) = sec^2 (u) - cos^2 (u)

    4)  \frac {1-cos (u)} { sin(u)} = \frac {1} {cosec(u) + cot(u)}

    Any help appreciated
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  2. #2
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    for number 2)
    left hand side
    = sec(u) + csc(u)cot(u)
    = 1/cos(u) + [1/sin(u)]*[cos(u)/sin(u)]
    = 1/cos(u) + [cos(u)/(sin(u))^2]
    =
    [(sin(u))^2 + (cos(u))^2]/[cos(u)*[sin(u)]^2]
    = 1/
    [cos(u)*[sin(u)]^2]
    = sec(u)*[csc(u)]^2
    = right hand side

    sorry about the bad syntax. hope it makes sense!
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  3. #3
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    Already solved these for you.
    Please don't double post.
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  4. #4
    Super Member angel.white's Avatar
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    Quote Originally Posted by jono View Post
    Prove -
    1)  sec (u) + tan (u) = \frac {1} {sec(u) - tan(u)}

    2)  sec (u) + cosec (u)cot(u) = sec(u)cosec^2(u)

    3)  sin^2 (u)(1+sec^2 (u) ) = sec^2 (u) - cos^2 (u)

    4)  \frac {1-cos (u)} { sin(u)} = \frac {1} {cosec(u) + cot(u)}

    Any help appreciated
    Okay, start with the Pythagorean theorem, which says a^{2}+b^{2}=c^{2}

    Now, sin(x) is the opposite side over the hypotenuse. And because the trigonometric functions are defined by the unit circle, we can set our triangle on the unit circle. Since the radius of the unit circle is 1, our hypotenuse will be 1. So sin(x) = opposite side over 1, so it is just the opposite side. This means the sin(x) gives you the y value of your triangle (because the length of the opposite side is how tall the triangle is). And cos(x) = adjacent length over hypotenuse, and thus is just the length of your triangle.

    So we see that we can plug these values into the Pythagorean theorem, length = cos(x), and height=sin(x)
    a^{2}+b^{2}=c^{2}\Rightarrow sin^{2}(x) + cos^{2}(x)=1
    (because 1 is the length of the radius on the unit circle, and that is our hypotenuse, and 1 squared is 1)

    So our equation is
    sin^{2}(x) + cos^{2}(x)=1

    1. Convert from trig functions to length ratios
    \frac{opp^{2}}{hyp^{2}} + \frac{adj^{2}}{hyp^{2}}=1

    2. Multiply by hyp^{2}
    opp^{2} + adj^{2}=hyp^{2}

    3. Divide by adj^{2}
    \frac{opp^{2}}{adj^{2}} + \frac{adj^{2}}{adj^{2}}=\frac{hyp^{2}}{adj^{2}}

    4. Convert back to trig functions
    tan^{2}(x) + 1=sec^{2}(x)

    5. Subtract tan^{2}
    1=sec^{2}(x)-tan^{2}(x)

    6. Because a^{2}-b^{2} = (a+b)(a-b)
    1=[sec(x) + tan(x)][sec(x)-tan(x)]

    7. And divide
    \frac{1}{sec(x)-tan(x)}=sec(x) + tan(x)
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