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Math Help - [SOLVED] Prove the identity, Stuck?

  1. #1
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    Exclamation [SOLVED] Prove the identity, Stuck?

    Hi, I tried doing this equation over and over, but I can't seem to get it! I've had sleepless nights for the past week...lol. So, please help me and put me out of my misery, I beg you...

    I need to prove that:
    \frac {1+sinA}{cosA} + \frac{cosA}{1+sinA} = 2secA

    this is what I came up with so far on the LHS, using 1=sinA+cosA:

    = \frac {(sinA+cosA)+sinA}{cosA} + \frac {cosA}{(sinA+cosA)+sinA}

    = \frac {sinA+cosA+sinA}{cosA} + \frac{cosA}{sinA+cosA+sinA}

    = \frac {(1-cosA)+cosA+(1-cosA)}{cosA} + \frac{cosA}{(1-cosA)+cosA+(1-cosA)}

    = \frac {2-cosA}{cosA} + \frac{cosA}{2-cosA}

    And that's where I got stuck. I don't even think my working out is right...

    PS. I tried using Wolfram|Alpha to solve it but it just tells me that the equation is true and gives me alternate forms, it doesn't show me steps which I need in order to fully understand, ya know?
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by mooniq View Post
    Hi, I tried doing this equation over and over, but I can't seem to get it! I've had sleepless nights for the past week...lol. So, please help me and put me out of my misery, I beg you...

    I need to prove that:
    \frac {1+sinA}{cosA} + \frac{cosA}{1+sinA} = 2secA

    this is what I came up with so far on the LHS, using 1=sinA+cosA:

    = \frac {(sinA+cosA)+sinA}{cosA} + \frac {cosA}{(sinA+cosA)+sinA}

    = \frac {sinA+cosA+sinA}{cosA} + \frac{cosA}{sinA+cosA+sinA}

    = \frac {(1-cosA)+cosA+(1-cosA)}{cosA} + \frac{cosA}{(1-cosA)+cosA+(1-cosA)}

    = \frac {2-cosA}{cosA} + \frac{cosA}{2-cosA}

    And that's where I got stuck. I don't even think my working out is right...

    PS. I tried using Wolfram|Alpha to solve it but it just tells me that the equation is true and gives me alternate forms, it doesn't show me steps which I need in order to fully understand, ya know?
    \begin{array}{lcl}<br />
\displaystyle \frac {1+\sin A}{\cos A} + \frac{\cos A}{1+\sin A} &=& \displaystyle \frac{(1+\sin A)^2+\cos^2 A}{\cos A\cdot (1+\sin A)}\\<br />
&=& \displaystyle \frac{2(1+\sin A)}{\cos A\cdot (1+\sin A)}\\<br />
&=& \displaystyle \frac{2}{\cos A}\\<br />
&=& 2\sec A<br />
\end{array}<br />
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  3. #3
    Super Member malaygoel's Avatar
    Joined
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    India
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    you are using the wrong identity

    Correct:
    1-sin^2A+cos^2A

    instead, you are using
    1=sinA+cosA
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  4. #4
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    thanks so much for the speedy response guys! I really appreciate it.
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