Trigonometry identities

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  • #1
chwala
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Homework Statement


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## {(tan φ+sec φ-1)/(tan φ-sec φ+1)}≡ {(1+sin φ)/cos φ}##[/B]


Homework Equations




The Attempt at a Solution


## (sin φ+1-cos φ)/(sin φ+cos φ-1)##[/B]
 

Answers and Replies

  • #2
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## (sin φ+1-cos φ)/(sin φ+cos φ-1)##
How did you get this?
It is impossible to tell where you need help if you don't show the steps you took so far.
 
  • #3
fresh_42
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To deal with expressions ##a-1## in a denominator, it is often useful to expand the quotient by ##a+1\,.##
 
  • #4
chwala
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To deal with expressions ##a-1## in a denominator, it is often useful to expand the quotient by ##a+1\,.##
That is what i did....let me look at my working again...
 
  • #5
fresh_42
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No, you expanded the entire thing by ##\cos \varphi##. Now you can go ahead and write the denominator as ##a-1## and make ##a^2-1## out of it.
 
  • #6
chwala
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No, you expanded the entire thing by ##\cos \varphi##. Now you can go ahead and write the denominator as ##a-1## and make ##a^2-1## out of it.
still not getting...how can you get ##a-1## in denominator?
look at my work now
##{(tan ψ+sec ψ-1)(tan ψ-sec ψ-1)}/{(tan ψ-sec ψ+1)(tan ψ-secψ-1)}##
let ## b= sec ψ-1##
we have
##{(tan ψ+b)(tanψ+b)}/{(tan^2ψ-b^2ψ)}##
after cancelling## (tan ψ+b)##
we get the original problem again!
 
  • #7
chwala
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No, you expanded the entire thing by ##\cos \varphi##. Now you can go ahead and write the denominator as ##a-1## and make ##a^2-1## out of it.
not really...how ## cos ψ?##
 
  • #8
chwala
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anyway, let me try do it and post my solution, i believe i am capable...then you can post alternative way of doing it..
 
  • #9
symbolipoint
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The lefthand member uses tangents and secants. The righthand member uses sines and cosines. Try starting with definition of tangent and secant on the left side; and see what other simplifications and algebraic steps you can pick...
 
  • #10
chwala
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I nailed it, i guess sometimes i am just too tired or not motivated. Here
## {(sin ψ+1-cosψ)/cosψ}.{(cos φ)/sin φ-1+cosφ)}##
##{(sin φ+1-cosφ)cos φ)/(sin φ-1+cosφ)cos φ)}##
##{sin ψcosψ+cosψ-cos^2ψ)/(sin φ-1+cosφ)cos φ)}##
##{sin ψcosψ+cosψ-1+sin^2ψ)/(sin φ-1+cosφ)cos φ)}##
##{(sin^2ψ-1+cosψ+sinψcosψ)/(sin φ-1+cosφ)cos φ)}##
##{(sin φ+1)(sinφ-1)+cosφ(1+sinφ)/(sin φ-1+cosφ)cos φ)}##
##{[(sin φ+1)][(sinφ-1+cosφ)]/(sin φ-1+cosφ)cos φ)}##
##(sin φ+1)/cosφ##
 
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  • #11
LCKurtz
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Your equations would look much nicer if you used \sin and \cos in your tex expressions.
For example, ##\sin\psi## versus ##sin\psi##.
 
  • #12
chwala
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There are 2 other methods to this....from my colleagues, i can share...
 
  • #13
fresh_42
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I have already mentioned one: expand the quotients by ##\cos \varphi + \sin \varphi +1##.
 

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