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Math Help - more basic trig

  1. #1
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    more basic trig

    prove the following trig identity:

    (1+sin theta/ cos theta) + (cos theta/ 1+sin theta) = 2/cos theta

    Thank you!
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  2. #2
    Member Henderson's Avatar
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    Multiply the top & bottom of your second fraction by 1-sin(\theta). Your denominator will then be 1-sin^2(\theta), which equals cos^2(\theta).

    Does that help you get started?
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  3. #3
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    \frac{1 + \sin x}{\cos x} + \frac{\cos x}{1 + \sin x} =

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

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

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

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

    \frac{2 \cos x}{\cos^2 x} =

    \frac{2}{\cos x}
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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by ss103 View Post
    prove the following trig identity:

    (1+sin theta/ cos theta) + (cos theta/ 1+sin theta) = 2/cos theta

    Thank you!
    Working on the left hand side:

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

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

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

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

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

    \frac{2}{\cos\theta}
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