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Math Help - complete the following identities.

  1. #1
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    complete the following identities.

    a stands for alpha b stands for beta

    a) cos a= b) 1+cot squared a=
    sin a


    c) tan(-a)= d) csc(90degrees - theta)=


    e) tan a - tan b =
    1+tan a tan b
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  2. #2
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    Quote Originally Posted by Dave19 View Post
    a stands for alpha b stands for beta

    a) cos a= b) 1+cot squared a=
    sin a


    c) tan(-a)= d) csc(90degrees - theta)=


    e) tan a - tan b =
    1+tan a tan b
    In order to prove an identity one must have both sides of the identity...

    If a) is
    \frac{cos(a)}{sin(a)}
    then you should know this. Hint: What are the definitions of tan(x), cot(x), sec(x), and csc(x)?

    b) This is one side of a "standard" identity. Look at the link in your other thread and you should find it.

    c) This is not so much of an identity as it is a simplification.
    tan(-a) = \frac{sin(-a)}{cos(-a)}
    So what are sin(-a) and cos(-a)?

    d) csc(90 - \theta) = \frac{1}{sin(90 - \theta)}
    How do you find sin(90 - \theta)?

    e) Hint: What is the form for tan(a + b)? (See that link again.)

    -Dan
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by Dave19 View Post
    a stands for alpha b stands for beta

    a) cos a= b) 1+cot squared a=
    sin a


    c) tan(-a)= d) csc(90degrees - theta)=


    e) tan a - tan b =
    1+tan a tan b
    a) \frac{\cos\alpha}{\sin\alpha}=\cot\alpha

    b) 1+\cot^2\alpha=\csc^2\alpha

    c) \tan(-\alpha)=-\tan\alpha

    d) \csc(90-\theta)=\sec\theta

    e) \frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}=\tan(\alpha-\beta)
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