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Solving Identites

  • Thread starter ritagogna
  • Start date
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1. Homework Statement

hey i really need some ones help i need to prove these 2 identities. im in grade 12 and ive been working on these questions for almost 2 hours and keep failing plzzz help
b. 1/1+sinx = sec2x - tanx/cosx


2. Homework Equations

-secx=1/cosx
-tanx=sinx/cosx
-sin2x+cos2x=1
-cscx=1/sinx

3. The Attempt at a Solution
So i tried proving the identity and got stuck...
i picked one side which was 1/1+sinx
then i got the conjugate
=1/1+sinx(1-sinx/1-sinx)
=1-sinx(1/sinx)
and this is where i have no idea what to do because i need to make it equal to sec2x-tanx/cosx
 

Answers and Replies

cristo
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By sec2x do you mean sec2x? If so, the equation is [tex]\frac{1}{1+sinx}=sec^2x-\frac{tanx}{cosx}[/tex].

Now, my advice would be to write out tanx and secx in terms of cosx and sinx. Then multiply the equation to get rid of the denominators. Try this and post what you get.
 
Tom Mattson
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=1/1+sinx(1-sinx/1-sinx)
I take it that you mean:

[tex]\frac{1}{1+\sin(x)}\frac{1-\sin(x)}{1-\sin(x)}[/tex]

=1-sinx(1/sinx)
I'm not sure of how you got this (or even of how to read it!), but it definitely looks wrong. You should multiply across the top and across the bottom in my expression above. Try that, and post what you get.
 
6
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so i worked it out and im pretty sure its 1- sinx/ cosx
 
Tom Mattson
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Nope, the cosine should be squared.
 
6
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no the equation is exacly what cristo got
 
6
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ok so its 1-sinx over cos2x
 
Tom Mattson
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no the equation is exacly what cristo got
No, I'm talking about the part of your solution that I quoted.
 
6
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wait no its 1/cos2x-sinx/cosx
 
Tom Mattson
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ok so its 1-sinx over cos2x
Do you mean cos2(x)? If so, then you're right. But if you mean cos(2x), then you're not right.
 
6
0
no i mean cos squared 2 x but i still havent solved the identity and i dont know where to go from here because ill just be making the equation more complicated by converting the identities
 
Tom Mattson
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no i mean cos squared 2 x
That's wrong. When you multiply [itex]1+\sin(x)[/itex] by [itex]1-\sin(x)[/itex], you get [itex]1-\sin^2(x)=\cos^2(x)[/itex]. The argument of the cosine is not [itex]2x[/itex].
 

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