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Math Help - verifying identities help needed

  1. #1
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    verifying identities help needed

    i have no idea! at all how to do it. plz help

    1) Cos X/Sec X - 1 - Cos X/tan^2 X = Cot^2 X

    2) Sin X / 1 - Cos X + 1 - Cos X / Sin X = 2CSC X
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  2. #2
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    first identity

    \frac{\cos\theta}{\sec\theta-1}-\frac{\cos\theta}{\tan^2\theta}=\cot^2\theta

    \frac{\cos\theta}{\frac{1}{\cos\theta}-1}-\frac{\cos\theta}{\frac{\sin^2\theta}{\cos^2\theta  }}

    \frac{\cos^2\theta}{1-\cos\theta}-\frac{cos^3\theta}{\sin^2\theta}

    \frac{\cos^2\theta}{1-\cos\theta}-\frac{cos^3\theta}{\sin^2\theta}

    \frac{\cos^2\theta}{1-\cos\theta}-\frac{cos^3\theta}{\sin^2\theta}

    \frac{\cos^2\theta}{1-\cos\theta}\cdot\frac{1+\cos\theta}{1+\cos\theta}-\frac{cos^3\theta}{\sin^2\theta}

    \frac{\cos^2\theta}{\sin^2\theta}=\cot^2\theta
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  3. #3
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    second identity

    \frac{\sin\theta}{1-\cos\theta}+\frac{1-\cos\theta}{\sin\theta}=2\csc\theta

    \frac{\sin\theta(1+\cos\theta)}{1-\cos^2\theta}+\frac{(1-\cos\theta)(\sin\theta)}{\sin^2\theta}

    \frac{2\sin\theta}{\sin^2\theta}

    \frac{2}{\sin\theta}=2\csc\theta
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  4. #4
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    Hello everyone
    Quote Originally Posted by satishgaire View Post
    ...1) Cos X/Sec X - 1 - Cos X/tan^2 X = Cot^2 X...
    Here's a little tip: where possible try to factorise an expression if you can, rather than expanding it. You'll usually keep the working much simpler that way.

    Now if you don't know that
    \sec^2\theta = 1 + \tan^2\theta
    then you should learn it. (If you can't see where it comes from, just divide both sides of \cos^2\theta+\sin^2\theta = 1 by \cos^2\theta.)

    Well, that means that:
    \tan^2\theta = \sec^2\theta - 1
    =(\sec\theta -1)(\sec\theta +1)...(1)
    Can you see that this means that we can factorise the original expression? Like this:
    \frac{\cos\theta}{\sec\theta-1} -\frac{\cos\theta}{\tan^2\theta}=\frac{\cos\theta}{  \sec\theta-1} -\frac{\cos\theta}{(\sec\theta -1)(\sec\theta +1)}, using (1)
    =\left(\frac{\cos\theta}{\sec\theta-1}\right)\left(1 -\frac{1}{\sec\theta +1}\right)

    =\left(\frac{\cos\theta}{\sec\theta-1}\right)\left(\frac{\sec\theta}{\sec\theta +1}\right)

    =\frac{1}{\tan^2\theta}, using (1) again.

    =\cot^2\theta
    Grandad
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