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Math Help - Simplify the Trig Expression...

  1. #1
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    Simplify the Trig Expression...

    I'm stuck and i don't know how to do this.


    (4tan^2(θ)cot^2(θ)/csc^2(θ))+(4/tan^2(θ)+1)
    Last edited by deezeejoey; February 6th 2008 at 11:23 AM.
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  2. #2
    GAMMA Mathematics
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    Quote Originally Posted by deezeejoey View Post
    I'm stuck and i don't know how to do this.


    (4tan^2(θ)cot^2(θ)/csc^2(θ))+(4/tan^2(θ)+1)
    Did you try reducing everything to sines and cosines?

    Hint: The second term would then be: \frac{4cos^2(\theta)}{sin^2(\theta)+cos^2(\theta)} = 4cos^2(\theta)
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  3. #3
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    Hello, deezeejoey!

    Do you remember any of the basic identities?


    \frac{4\tan^2\!\theta\cot^2\!\theta}{\csc^2\!\thet  a}+ \frac{4}{\tan^2\!\theta+1}

    \text{The first fraction is: }\;\frac{4(\overbrace{\tan\theta\cot\theta}^{\text  {This is 1}})^2}{\csc^2\!\theta} \;=\;\frac{4}{\csc^2\!\theta} \;=\;4\sin^2\!\theta

    \text{The second fraction is: }\;\frac{4}{\underbrace{\tan^2\!\theta + 1}_{\text{This is }\sec^2\theta}} \;=\;\frac{4}{\sec^2\!\theta} \;=\;4\cos^2\!\theta


    \text{The problem becomes: }\;4\sin^2\!\theta + 4\cos^2\!\theta \;=\;4\underbrace{(\sin^2\!\theta + \cos^2\!\theta)}_{\text{This is 1}} \;=\;4\cdot1 \;=\;4

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