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Math Help - Trigonometric Identities cosecA

  1. #1
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    Question Trigonometric Identities cosecA

    (Extract from Cambridge O-Level Additional Mathematics June 2012 exam paper 1, question 3)


    Show that

    cotA+\frac{sinA}{1+cosA}\equiv cosecA


    I can't get rid of the 1+cosA when simplifying RHS, it keeps getting complex.
    Help please.
    Last edited by zikcau25; November 6th 2012 at 05:51 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: Trigonometric Identities cosecA

    Quote Originally Posted by zikcau25 View Post
    Show that
    cotA+\frac{sinA}{1+cosA}\equiv cosecA


    I can't get rid of the 1+cosA when simplifying RHS, it keeps getting complex.
    Help please.
    There can be easier solutions to the problem than the way I do it, but usually it works out not too bad.

    I always replace things like cot(A) and cosec(A) (and whatever else) with sines and cosines. So for your identity we have
    \frac{cos(A)}{sin(A)} + \frac{sin(A)}{1 + cos(A)} = \frac{1}{sin(A)}

    Generally the next step is to add fractions. So we have fractions...let's add them. The common denominator will simply be sin(A) \cdot (1 + cos(A))}

    So what's the LHS look like?

    -Dan
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    MHF Contributor MarkFL's Avatar
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    Re: Trigonometric Identities cosecA

    Another approach:

    Try to derive the following:

    \tan\left(\frac{A}{2} \right)=\csc(A)-\cot(A)

    \tan\left(\frac{A}{2} \right)=\frac{\sin(A)}{1+\cos(A)}
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  4. #4
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    Re: Trigonometric Identities cosecA

    Hello, zikcau25!

    \text{Show that : }\:\cot A+\frac{\sin A}{1+\cos A}\:=\:\csc A

    We have: . \frac{\cos A}{\sin A} + \frac{\sin A}{1 + \cos A} \;=\;\frac{\cos A}{\sin A} + \frac{\sin A}{1 + \cos A}\cdot\color{blue}{\frac{1-\cos A}{1-\cos A}}

    . . . . . . =\;\frac{\cos A}{\sin A} + \frac{\sin A(1-\cos A)}{1-\cos^2\!A} \;=\;\frac{\cos A}{\sin A} + \frac{\sin A(1-\cos A)}{\sin^2\!A}

    . . . . . . =\;\frac{\cos A}{\sin A} + \frac{1-\cos A}{\sin A} \;=\;\frac{\cos A + 1 - \cos A}{\sin A} \;=\;\frac{1}{\sin A} \;=\;\csc A
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