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Core 3- Trig

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    [sec(x) - 1 ]/tan(x) = [ tan (x)]/ sec(x) +1

    So, i have to prove this identity is true. Ive managed to show it by dividing both sides by one of the sides, however i cant think how to start from either the left or right, and work myself to the other side. Any help would be awesome
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    \frac{sec-1}{tan} = [\frac{1}{cos} - 1]\frac{cos}{sin} = \frac{1-cos}{sin} = \frac{1-cos^2}{sin(1+cos)} = \frac{sin}{1+cos} = \frac{tan}{sec+1}
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    (Original post by antonio108)
    [sec(x) - 1 ]/tan(x) = [ tan (x)]/ sec(x) +1

    So, i have to prove this identity is true. Ive managed to show it by dividing both sides by one of the sides, however i cant think how to start from either the left or right, and work myself to the other side. Any help would be awesome
     \displaystyle \frac{secx-1}{tanx} \equiv \frac{tanx}{secx+1}

    LHS =
     \displaystyle \frac{secx-1}{tanx} = \frac{\frac1{cosx}-1}{\frac{sinx}{cosx}} = \frac{\frac{1-cosx}{cosx}}{\frac{sinx}{cosx}}=  \frac{1-cosx}{sinx}=\frac{(1-cosx)(1+cosx)}{sinx(1+cosx)}}
     \displaystyle \frac{sin^2x}{sinx(1+cosx)} = \frac{sinx}{1+cosx}=\frac{\frac{  sinx}{cosx}}{\frac{1+cosx}{cosx}  }=\frac{tanx}{secx+1}
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    Several methods, here is mine.




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    Ah thank you very much, just looking through and slowly realising what i have to look out for.
    Again, thanks!

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