Identities sin, cos, tan etc. stuff

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SUMMARY

The discussion focuses on simplifying the expression \(\frac{\cos x}{1+\sin x} + \frac{1+\sin x}{\cos x}\). Participants detail their attempts to manipulate the equation by multiplying fractions and expanding numerators. The correct simplification leads to the conclusion that the expression equals \(2 \sec x\). Key steps include expanding \((1+\sin x)^2\) and using the identity \(\cos^2 x = 1 - \sin^2 x\) for further simplification.

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UltimateSomni
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Homework Statement



((cos x)/(1+sin x))+((1+sin x)/(cos x))

Homework Equations

2012-03-05_12-13-36_59.jpg

The Attempt at a Solution



multiplied the left equation by (cos x)/(cos x) and the right fraction by (1+sin x)/(1+sin x)

get ((cosx)^2 +(1+sinx)^2) / (1+sinx) (cos x)

and I have no idea where to go from here
 
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UltimateSomni said:

Homework Statement



((cos x)/(1+sin x))+((1+sin x)/(cos x))
What are you trying to do, simplify the expression above?
UltimateSomni said:

Homework Equations




2012-03-05_12-13-36_59.jpg



The Attempt at a Solution



multiplied the left equation by (cos x)/(cos x) and the right fraction by (1+sin x)/(1+sin x)

get ((cosx)^2 +(1+sinx)^2) / (1+sinx) (cos x)

and I have no idea where to go from here
 
Yes, the answer being 2 sec x
 
Expand the numerator and simplify.
((cosx)^2 +(1+sinx)^2) / (1+sinx) (cos x)
 
(1-sinx^2) +1 + sinx +1 - sin x

(1- sinx^2) +2 / (1+sin x)(cos x)

I don't see how that helps
 
Mark44 said:
Expand the numerator and simplify.
((cosx)^2 +(1+sinx)^2) / (1+sinx) (cos x)

UltimateSomni said:
(1-sinx^2) +1 + sinx +1 - sin x
The only thing you did that makes sense is replacing cos2(x) with 1 - sin2(x). I don't get what you did go go from (1 + sin(x))2 to 1 + sin(x) + 1 - sin(x).

Mark44 said:
(1- sinx^2) +2 / (1+sin x)(cos x)

I don't see how that helps
 
Then what simplifying is there to do?
 
Apparently I wasn't clear. (1 + sin(x))2 ≠ 1 + sin(x) + 1 - sin(x), which seems to be what you're saying.

The "simplifying" that you need to do is to expand (1 + sin(x))2 to something it is actually equal to.
 
Okay it equals (1+sinx)(1-sinx)
still not way to get to 2secx
 
  • #10
Because that's wrong, too. (1 + sin(x))2 ≠ (1+sinx)(1-sinx), if that's what you're saying.

How do you expand (1 + x)2? This is a similar kind of problem.
 
  • #11
Okay it equals (1+sinx)(1+sinx)
still no way to get to 2secx
 
  • #12
UltimateSomni said:
Okay it equals (1+sinx)(1+sinx)

Expand the expression: apply distributivity to work out the brackets.
 
  • #13
All right got it
 

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