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anonymous

  • 5 years ago

I'm working on trig identities and in need of some help : (1-sec x)/(1+sec x)= (cos x-1)/(cos x+1) AND (cos x)/(1+sin x) + (1+sin x)/ (cos x)= 2 sec x

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  1. anonymous
    • 5 years ago
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    Your first one can be solved by the definition of secant:\[\frac{1-\sec x}{1+ \sec x}=\frac{1-\frac{1}{\cos x}}{1+\frac{1}{\cos x}}\]

  2. anonymous
    • 5 years ago
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    \[=\frac{\frac{\cos x-1}{\cos x}}{\frac{\cos x +1}{\cos x}}=\frac{\cos x -1}{\cos x +1}\]

  3. anonymous
    • 5 years ago
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    (1-1/cosx)/(1+1/cosx) =(cosx-1)/cosx/(cosx+1)/cosx =(cosx-1)(cosx+1)

  4. anonymous
    • 5 years ago
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    \[\frac{\cos x}{1+\sin x}+\frac{1+ \sin x }{\cos x}=\frac{\cos^2x+(1+\sin x)^2}{\cos x (1+\sin x)}\]

  5. anonymous
    • 5 years ago
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    \[=\frac{\cos^x+1+2\sin x + \sin^2x}{\cos x(1+\sin x)}=\frac{2+2\sin x}{\cos x (1+\sin x)}\]

  6. anonymous
    • 5 years ago
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    \[=\frac{2(1+\sin x)}{\cos x (1+\sin x)}\]

  7. anonymous
    • 5 years ago
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    \[=\frac{2}{\cos x}=2\sec x\]

  8. anonymous
    • 5 years ago
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    What are the identities used to get from this one fraction to the next? cos x +1+2sin x+ sin ^{2} x div cos x times 1 + sin x = 2+ 2 sin x div cos x times 1 + sin x

  9. anonymous
    • 5 years ago
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    I thought that's what you were writing about...\[\frac{\cos^2x+1+2\sin x+\sin^2x}{\cos x(1+\sin x)}\]\[=\frac{(\cos^2 x+ \sin^2 x)+1+2\sin x}{\cos x (1+\sin x)}\]\[=\frac{1+1+2\sin x}{\cos x(1+\sin x)}=\frac{2+2\sin x}{\cos x(1 + \sin x)}=\frac{2(1+\sin x)}{\cos x(1+\sin x)}\]

  10. anonymous
    • 5 years ago
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    It's just the \[\cos^2x + \sin^2 x =1\] identity.

  11. anonymous
    • 5 years ago
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    Both of you, thank you so very much for your expertise with this problems, it helps me greatly!!!! I really appreciate it! :)

  12. anonymous
    • 5 years ago
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    You're welcome :) Thanks for the appreciation - not everyone bothers to say thank you!

  13. anonymous
    • 5 years ago
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    I mean it, I am very greatful for the help, thanks for taking the time to help me, you deserve a thank you, you both do :)

  14. anonymous
    • 5 years ago
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    my pleasure;-) @lokisan i bother if my network doesnot fail;-)

  15. anonymous
    • 5 years ago
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    my pleasure;-) @lokisan i bother if my network doesnot fail;-)

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