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Math Help - tan cot sec

  1. #1
    Senior Member furor celtica's Avatar
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    Angry tan cot sec

    solve this equation for A in the range 0<A<360
    tanA + cotA = 2secA
    i found A= 30, 150, 210, 330 degrees and all of these results fit the tangent function i solved for. but the answer includes only A = 30 and 150 degrees. why is this?
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  2. #2
    MHF Contributor
    Prove It's Avatar
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    Quote Originally Posted by furor celtica View Post
    solve this equation for A in the range 0<A<360
    tanA + cotA = 2secA
    i found A= 30, 150, 210, 330 degrees and all of these results fit the tangent function i solved for. but the answer includes only A = 30 and 150 degrees. why is this?
    \tan{A} + \cot{A} = 2\sec{A}

    \frac{\sin{A}}{\cos{A}}+ \frac{\cos{A}}{\sin{A}} = \frac{2}{\cos{A}}

    \frac{\sin^2{A}}{\sin{A}\cos{A}} + \frac{\cos^2{A}}{\sin{A}\cos{A}} = \frac{2\sin{A}}{\sin{A}\cos{A}}

    \frac{\sin^2{A} + \cos^2{A}}{\sin{A}\cos{A}} = \frac{2\sin{A}}{\sin{A}\cos{A}}

    \sin^2{A} + \cos^2{A} = 2\sin{A}

    1 = 2\sin{A}

    \sin{A} = \frac{1}{2}

    Since \sin is only positive in the first and second quadrants, in the domain 0^{\circ} \leq x \leq 360^{\circ}, there will only be two solutions.

    A = 30^{\circ} or A = 150^{\circ}.
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  3. #3
    MHF Contributor
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    Hello furor celtica
    Quote Originally Posted by furor celtica View Post
    solve this equation for A in the range 0<A<360
    tanA + cotA = 2secA
    i found A= 30, 150, 210, 330 degrees and all of these results fit the tangent function i solved for. but the answer includes only A = 30 and 150 degrees. why is this?
    ProveIt has given you the correct solution. But to try to answer your question, did you, in your solution, square both sides of the equation at some point? If you did, that will be why you got the extra, unwanted solutions.

    You can see a similar thing if you square both sides in a simple non-trig equation. For example:

    x+1=4

    The solution is, of course, x = 3. But if you square both sides:

    (x+1)^2 = 16

    \Rightarrow x^2 + 2x + 1 = 16

    \Rightarrow x^2 +2x - 15=0

    \Rightarrow (x +5)(x-3)=0

    \Rightarrow x = -5 or 3

    And the value x = -5 doesn't satisfy x+1 = 4, but x+1 = -4. But of couse, both 4^2 and (-4)^2 give the value 16.

    So, always beware of squaring both sides of an equation: you'll sometimes introduce extra, unwanted solutions if you do.

    Grandad
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  4. #4
    Senior Member furor celtica's Avatar
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    thank you both, i will keep that in mind, i did effectively square the equation somewhere in my reasoning.
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