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Math Help - Prove!

  1. #1
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    Prove!

    Hi,


    Please help.


    Prove that:
    (sin θ-kos θ+1)/(sin θ+kos θ-1)= (sin θ+1)/(kos θ)


    (kos θ/ 1-tan x) + (sin θ/1-kot θ) = sin θ + kos θ


    Solve that tanx = π - 2x , 0<x<π/2


    Thnks.
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  2. #2
    MHF Contributor red_dog's Avatar
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    Re: Prove!

    1) \frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}  +2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2} \cos\frac{\theta}{2}-2\sin^2\frac{\theta}{2}}=\frac{\sin\frac{\theta}{2  }+\cos\frac{\theta}{2}}{\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}=\\=\frac{\left(\sin \frac{\theta}{2}+\cos \frac{\theta}{2}\right)^2}{\cos^2 \frac{\theta}{2}-\sin^2 \frac{\theta}{2}}=\frac{1+\sin \theta}{\cos \theta}

    2) \frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta}=\frac{\cos^2 \theta}{\cos \theta-\sin \theta}+\frac{\sin^2 \theta}{\sin \theta-\cos \theta}=\\=\frac{\cos^2 \theta-\sin^2 \theta}{\cos \theta-\sin \theta}=\cos \theta+\sin \theta

    I don't understand the last part of the problem
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  3. #3
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    Re: Prove!

    Thank you. May I know how (sinθ/2 = cos θ/2)^2 = 1+sin θ ?this part I m not understanding.
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  4. #4
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    Re: Prove!

    Hi, I got it. Thank you very much for ur help! =D
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  5. #5
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    Re: Prove!

    Alternatively we can also get the solution as indicated
    Prove!-06-sep-13.png
    Thanks from Shayna
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  6. #6
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    Re: Prove!

    hi,

    SORRY, I can't get the steps at line 2. sec^2 x -tan
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  7. #7
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    Re: Prove!

    hi,

    SORRY, I can't get the steps at line 2. sec^2 x -tan^2 x -----> 1-(sec x -tan x) ? TQ.
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  8. #8
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    Re: Prove!

    It was (sec x +tan x ) -1 = ( sec x - tan x ) - ( sec^2 x - tan^2 x ) .... Because sec^2 x -tan^2 x = 1
    = ( sec x - tan x ) - (( sec x - tan x ) ( sec x +tan x ) )= ( sec x +tan x ) [ 1 - ( sec x - tan x ) ] etc
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