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Math Help - Trigonometric Identities

  1. #1
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    Trigonometric Identities

    Prove the following identity
    cosec (theta) + tan (theta) = 1 / [cosec (theta) - tan (theta)]
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  2. #2
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    Re: Trigonometric Identities

    Quote Originally Posted by Ilsa View Post
    Prove the following identity
    cosec (theta) + tan (theta) = 1 / [cosec (theta) - tan (theta)]
    What have you tried?

    Where are you stuck?

    Work on the left side. Write cosecant and tangent functions in terms of sine and cosine. Combine into one rational expression.
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  3. #3
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    Re: Trigonometric Identities

    cosec (theta) + tan (theta)
    = 1 / sin (theta) + (sin (theta)/cos (theta))
    = [cos (theta) + sin ^2 (theta) ] / sin (theta) cos (theta)
    I require help for further steps
    If I substitute sin^2 (theta) = 1 - cos^2 (theta)
    then:
    = [cos (theta) + 1 - cos^2 (theta) ] / sin (theta) cos (theta)
    With the above expression I am unable to prove the identity.
    This is where I'm stuck.
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  4. #4
    Member sbhatnagar's Avatar
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    Exclamation Re: Trigonometric Identities

    Quote Originally Posted by Ilsa View Post
    Prove the following identity
    cosec (theta) + tan (theta) = 1 / [cosec (theta) - tan (theta)]
    Check your question. You cannot prove it because it is false.

    \csc(\theta)+\tan(\theta)\neq \frac{1}{\csc(\theta)-\tan(\theta)}

    If you don't believe me ,then look:

    \csc(\theta)+\tan(\theta)=\frac{1}{\csc(\theta)-\tan(\theta)}

    (\csc(\theta)+\tan(\theta))(\csc(\theta)-\tan(\theta))=1

    \csc^2(\theta)-\tan^2(\theta)=1

    \csc^2(\theta)=1+\tan^2(\theta)

    \csc^2(\theta)=\sec^2(\theta) (which is wrong!!)


    According to me your question should read : Prove \csc(\theta)+\cot(\theta)=\frac{1}{\csc(\theta)-\cot(\theta)}
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  5. #5
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    Re: Trigonometric Identities

    Hello, Ilsa!

    The statement is not true. .Is there a typo?


    \text{Prove: }\: \csc\theta + \cot\theta \:=\:\frac{1}{\csc\theta - \cot\theta} . This is true

    Multiply the right side by \frac{\csc\theta + \cot\theta}{\csc\theta + \cot\theta}

    . . \frac{1}{\csc\theta - \cot\theta} \cdot\frac{\csc\theta + \cot\theta}{\csc\theta + \cot\theta} \;\;=\;\;\frac{\csc\theta + \cot\theta}{\underbrace{\csc^2\!\theta - \cot^2\!\theta}_{\text{This is 1}}} \;\;=\;\; \csc\theta + \cot\theta
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