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Question which the teacher cant help with

  1. Nov 24, 2005 #1
    a)state the value of sec^2x-tan^2x
    b) the angle A is such that secA + tanA = 2. show that secA-tanA=1/2, and hence find the exact value of cos A.

    part a was a doddle so to speak, however, part b was on the contrary; it took our class half an hour, but to avail

    if someone could shed light on this problem, i would be much obliged

    thank you in advance
     
  2. jcsd
  3. Nov 24, 2005 #2

    HallsofIvy

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    Not all that difficult, surely.
    [tex] sec A= \frac{1}{cos A}[/tex] and [tex]tan A\frac{sin A}{cos A}[/tex]
    so [tex]sec A+ tan A= \frac{1+ sin A}{cos A}= 2[/tex]
    Looks to me like an obvious thing to try is to multiply numerator and denominator of that equation by 1- sin A:
    [tex]\frac{(1+ sin A)(1- sin A)}{(cos A)(1- sin A)}= 2[/tex]
    [tex]\frac{1- sin^2 A}{(cos A)(1- sin A)}= \frac{cos^2 A}{(cos A)(1- sin A)}= \frac{cos A}{1- sin A}= 2[/tex]
    Then
    [tex] \frac{1- sin A}{cos A}= \frac{1}{2}[/tex]
    [tex] \frac{1}{cos A}- \frac{sin A}{cos A}= sec A- tan A= \frac{1}{2}[/tex].
    Add the two equations to eliminate tan A and then invert.
     
  4. Nov 24, 2005 #3
    thanks. i think we were all just looking at it form the wron perspective. much obliged
     
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