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Math Help - Trigo Qn

  1. #1
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    Trigo Qn

    Hi i need help for this question. Stucked at it for a long time

    Given That Sec A - tan A = 2. determine the value of ( sec A + tan A ) without
    solving for A.
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  2. #2
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    Quote Originally Posted by Nuts View Post
    Hi i need help for this question. Stucked at it for a long time

    Given That Sec A - tan A = 2. determine the value of ( sec A + tan A ) without
    solving for A.
    On the one hand, (\sec (A) - \tan (A))(\sec (A) + \tan (A)) = \sec^2 (A) - \tan^2 (A) = 1 (this follows from the basic Pythagorean Identity).

    On the other hand, it's given that (\sec (A) - \tan (A))(\sec (A) + \tan (A)) = 2 (\sec (A) + \tan (A)).

    Therefore .....
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  3. #3
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    Quote Originally Posted by Nuts View Post
    Hi i need help for this question. Stucked at it for a long time

    Given That Sec A - tan A = 2. determine the value of ( sec A + tan A ) without
    solving for A.
    \displaystyle\frac{1}{CosA}-\frac{SinA}{CosA}=\frac{1-SinA}{CosA}=2\Rightarrow\ 1-SinA=2CosA

    We want to find

    \displaystyle\frac{1+SinA}{CosA}

    \displaystyle\ (1-SinA)(1+SinA)=2CosA(1+SinA)

    1-Sin^2A=Cos^2A=2CosA(1+SinA)\Rightarrow\ CosA=2(1+SinA)

    from which follows the result.
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    On the one hand, (\sec (A) - \tan (A))(\sec (A) + \tan (A)) = \sec^2 (A) - \tan^2 (A) = 1 (this follows from the basic Pythagorean Identity).

    On the other hand, it's given that (\sec (A) - \tan (A))(\sec (A) + \tan (A)) = 2 (\sec (A) + \tan (A)).

    Therefore .....
    Hmm i do not understand how does Sec A - tan A = 2 changes to (\sec (A) - \tan (A))(\sec (A) + \tan (A)) = 2 (\sec (A) + \tan (A)) . Could you explain
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  5. #5
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    Usually the easiest way to simplify trigonometric expressions is to multiply by the conjugate, then apply a trigonometric identity.

    In your case, you are told

    \sec{A} - \tan{A} = 2

    (\sec{A} - \tan{A})(\sec{A} + \tan{A}) = 2(\sec{A} + \tan{A})

    \sec^2{A} - \tan^2{A} = 2(\sec{A} + \tan{A}).


    Now use the Pythagorean Identity 1 + \tan^2{\theta} \equiv \sec^2{\theta} to simplify the LHS and you should be able to solve for \sec{A} + \tan{A}.
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  6. #6
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    Quote Originally Posted by Archie Meade View Post
    \displaystyle\frac{1}{CosA}-\frac{SinA}{CosA}=\frac{1-SinA}{CosA}=2\Rightarrow\ 1-SinA=2CosA

    We want to find

    \displaystyle\frac{1+SinA}{CosA}

    \displaystyle\ (1-SinA)(1+SinA)=2CosA(1+SinA)

    1-Sin^2A=Cos^2A=2CosA(1+SinA)\Rightarrow\ CosA=2(1+SinA)

    from which follows the result.
    Ok i solved it thanks!
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  7. #7
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    Quote Originally Posted by Nuts View Post
    Ok i solved it thanks!
    I thought you might have tried it using Sine and Cosine,
    so I posted in that way.

    However, it's best to know the technique shown by mr fantastic and Prove It.
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