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 Post subject: Solving trig identitiy
PostPosted: Mon, 14 May 2012 03:02:01 UTC 
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Hello. I am trying to solve the trig identity

1/(1-sinx) - 1/(1+sinx) = 2 secx * tanx

I started on the right hand side, finding

2/cos(x) * sin(x)/cos(x) combined the like terms to find,

2sin(x) / cos(x)^2 which was further broken into

2sin(x) / 1-sin(x)^2 and then expanded to

2sin(x) / 1-sin(x) * 1+sin(x)

Any ideas on what identities to use to continue solving the identity? Thanks.


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PostPosted: Mon, 14 May 2012 03:14:34 UTC 
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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supersam1130 wrote:
Hello. I am trying to solve the trig identity

1/(1-sinx) - 1/(1+sinx) = 2 secx * tanx

I started on the right hand side, finding

2/cos(x) * sin(x)/cos(x) combined the like terms to find,

2sin(x) / cos(x)^2 which was further broken into

2sin(x) / 1-sin(x)^2 and then expanded to

2sin(x) / 1-sin(x) * 1+sin(x)

Any ideas on what identities to use to continue solving the identity? Thanks.


Note:

{1\over 1-x}+{1\over 1+x}=?

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PostPosted: Mon, 14 May 2012 03:21:59 UTC 
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So, the denominators have to be the same. 2 / (1 + x)* (1 - x). I have almost solved the equation, but I don't understand how to get the right side to be subtraction instead of addition. My math teacher is notorious for making errors. Is it possible he meant for the original left hand side to be an addition of the two fractions?


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PostPosted: Mon, 14 May 2012 03:25:32 UTC 
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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supersam1130 wrote:
So, the denominators have to be the same. 2 / (1 + x)* (1 - x). I have almost solved the equation, but I don't understand how to get the right side to be subtraction instead of addition. My math teacher is notorious for making errors. Is it possible he meant for the original left hand side to be an addition of the two fractions?


Well if you look at the mechanics of adding the two, what happens when you subtract them?

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