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Math Help - prove identity

  1. #1
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    prove identity

    Prove the following identites;

     cosecAsec^{2}A = cosecA +tanAsecA

    I know you have to use these identiests;  1 + tan^{2} A = sec^{2}A
     1+ cot^{2} A = cosec^{2}A
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  2. #2
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    Hello Tweety
    Quote Originally Posted by Tweety View Post
    Prove the following identites;

     cosecAsec^{2}A = cosecA +tanAsecA

    I know you have to use these identiests;  1 + tan^{2} A = sec^{2}A
     1+ cot^{2} A = cosec^{2}A
    \csc A \sec^2A = \csc A(1 + \tan^2 A)

    = \csc A + \csc A \tan^2A

    = \csc A + \frac{\sin^2 A}{\sin A\cos^2A}

    = \csc A + \frac{\sin A}{\cos A}\cdot\frac{1}{\cos A}

    =\csc A + \tan A \sec A

    Grandad
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  3. #3
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    Hello, Tweety!

    Sometimes it's easier to change everything to sines and cosines.


    Prove: . \csc\!A\sec^2\!\!A \:=\: \csc\! A +\tan\! A\sec\! A

    On the right side, we have: . \frac{1}{\sin\! A} + \frac{\sin\! A}{\cos\! A}\!\cdot\!\frac{1}{\cos\! A} \;\;=\;\;\frac{1}{\sin\! A} + \frac{\sin\! A}{\cos^2\!\!A}

    Get a common denominator: . \frac{1}{\sin\! A}\cdot{\color{blue}\frac{\cos^2\!\!A}{\cos^2\!\!A  }} + \frac{\sin\!A}{\cos^2\!\!A}\cdot{\color{blue}\frac  {\sin\!A}{\sin\!A}} \;\;=\;\;\frac{\cos^2\!\!A}{\sin\!A\cos^2\!\!A} + \frac{\sin^2\!\!A}{\sin\!A\cos^2\!\!A}

    . . = \;\frac{\overbrace{\cos^2\!\!A + \sin^2\!\!A}^{\text{This is 1}}}{\sin\!A\cos^2\!\!A} \;=\;\frac{1}{\sin\!A\cos^2\!\!A}\;=\;\frac{1}{\si  n\!A}\cdot\frac{1}{\cos^2\!\!A} \;=\;\csc\!A\sec^2\!\!A

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