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Math Help - Trig Identities

  1. #1
    Junior Member casey_k's Avatar
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    Trig Identities

    Not sure how to approach this problem:
    Attached Thumbnails Attached Thumbnails Trig Identities-trig-identities.jpg  
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  2. #2
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    \frac{\sin{x}+\tan{x}}{\cos{x}+1}

    Multiply by \frac{\frac{1}{\cos{x}}}{\frac{1}{\cos{x}}} to get:

    \frac{\tan{x}+\frac{\tan{x}}{\cos{x}}}{1+\frac{1}{  \cos{x}}}

    Take \tan{x} common factor, and we're done.
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  3. #3
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    Quote Originally Posted by casey_k View Post
    Not sure how to approach this problem:
    Actually, with your working, you could've taken sin(x) common factor, then you can see a cancellation.
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  4. #4
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    Hello, casey_k!

    Another approach . . .


    \frac{\sin x + \tan x}{\cos x + 1} \;=\;\tan x

    We have: . \frac{\sin x + \tan x}{\cos x + 1} \;=\;\frac{\sin x + \frac{\sin x}{\cos x}}{\cos x + 1} \;=\;\frac{\sin x(1 + \frac{1}{\cos x})}{\cos x + 1}<br />

    Multiply by \frac{\cos x}{\cos x}\!:\;\;\frac{\cos x\cdot\sin x(1 + \frac{1}{\cos x})}{\cos x\cdot(\cos x + 1)} \;=\;\frac{\sin x(\cos x + 1)}{\cos x(\cos x + 1)} \;=\;\frac{\sin x}{\cos x} \;=\;\tan x

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