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Math Help - Prove this identity

  1. #1
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    Prove this identity

    I'm not sure how to prove these expressions true...the question is:

    Prove that each expression is an identity for all values for which it is defined.

    <br />
 \frac {sin\theta} {sin\theta+cos\theta} = \frac {tan\theta} {1+tan\theta}<br />

    and,

    <br />
\frac {sin^2\theta+2cos\theta-1} {sin^2+3cos\theta-3} = \frac {1} {1-sec\theta}<br />

    Thank you!
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  2. #2
    MHF Contributor harish21's Avatar
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    \dfrac{\sin x}{\sin x+\cos x}

    divide the numerator and denominator by cosx

    \dfrac{\frac{\sin x}{\cos x}}{\frac{\sin x+\cos x}{\cos x}}


    For the other question, show some effort. Where exactly are you stuck?
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  3. #3
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    Oh, wow, that was so obvious hahaha
    Thanks!!

    Any ideas on the second one?
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  4. #4
    MHF Contributor harish21's Avatar
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    \dfrac{\sin^2 x+2\cos x-1}{\sin^2 x+3\cos x-3}

    = \dfrac{1-\cos^2 x+2\cos x-1}{1-\cos^2 x+3\cos x-3}

    = \dfrac{-\cos^2 x+2\cos x}{-\cos^2 x+3\cos x-2}

    =\dfrac{-\cos x(cos x-2)}{-(\cos^2 x-3\cos x+2)}

    =\dfrac{\cos x(cos x-2)}{(\cos^2 x-3\cos x+2)}

    factorize the denominator and change all cosx into secx..
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  5. #5
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    That makes sense now, thank you!
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