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Math Help - Trig Identity Help

  1. #1
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    Exclamation Trig Identity Help

    One question left and i got stuck on it, i am starting to get the hang of these things though.

    Thanks, joe


    sec x + csc x / 1 + tan x = cscx
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by DaCoo911 View Post
    One question left and i got stuck on it, i am starting to get the hang of these things though.

    Thanks, joe


    sec x + csc x / 1 + tan x = cscx
    Hi DaCoo911,

    Working with the left hand side,

    \dfrac{\sec x + \csc x}{1+\tan x}=

    \dfrac{\dfrac{1}{\cos x}+\dfrac{1}{\sin x}}{1+\dfrac{\sin x}{\cos x}}=

    \dfrac{\dfrac{\sin x+ \cos x}{\sin x \cos x}}{1+\dfrac{\sin x}{\cos x}}=

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

    \dfrac{\sin x+ \cos x}{\sin x \cos x}\cdot \dfrac{\cos x}{\sin x+\cos x}=

    \dfrac{1}{\sin x}=

    \csc x
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  3. #3
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    Dacoo,

    1. \frac{1}{cosx}=secx

    2. \frac{1}{sinx}=csecx

    3. \frac{sin}{cosx}=tgx

    Find common denominator and you'll have

    \frac{(sinx+cosx)/(cosxsinx)}{(cosx+sinx)/cosx}

    cancel cosx+cosx then you got \frac{cosx}{cosxsinx}, cancel cosinuses and finally you got 1/sinx which is second identity i wrote.
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