anonymous
  • anonymous
how do i verify (cos x/ 1 + sin x) + (1 + sin x/ cos x) = 2 sec x? i don't know how to verify anything at all can someone explain how it works?
Mathematics
chestercat
  • chestercat
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Nnesha
  • Nnesha
\[\huge\rm \frac{ \cos(x) }{ 1+\sin(x) } +\frac{ 1+\sin(x) }{ \cos(x) }\] 1)what is the common denominator ?
anonymous
  • anonymous
i dont know
anonymous
  • anonymous
i really dont know anything about this

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Nnesha
  • Nnesha
alright let's change that to `simple` algebra
Nnesha
  • Nnesha
\[\huge\rm \frac{ x }{ y }+\frac{ y }{ x }\] what's the common denominator ?
anonymous
  • anonymous
xy
Nnesha
  • Nnesha
okay let's say y represent 1+sinx and x = cos(x) \[y=1+\sin(x)\] \[x=\cos(x)\] now can you tell me what would be the common denominator\[\huge\rm \frac{ \cos(x) }{ 1+\sin(x) } +\frac{ 1+\sin(x) }{ \cos(x) }\] for this question :=)
anonymous
  • anonymous
1+sinx*cosx ?
Nnesha
  • Nnesha
PERFECT!
anonymous
  • anonymous
hi i speak Arabic
anonymous
  • anonymous
salam alakom
anonymous
  • anonymous
Nnesha
Nnesha
  • Nnesha
\[\huge\rm \frac{ \cos(x) }{ \color{red}{1+\sin(x)} } +\frac{ 1+\sin(x) }{\color{blue}{ \cos(x)} }\] \[\huge\rm \frac{ \cos(\color{blue}{cos}) + (\color{reD}{1+\sin})(1+\sin)}{ \cos(x) (1+\sin(x))}\] multiply the numerator of `1st fraction` by the `denominator` of 2nd fraction multiply the numerator of `2nd fraction` with the denominator of `1st fraction`
anonymous
  • anonymous
so would it become cos^2x and (1+sin)^2 on the numerator??
Nnesha
  • Nnesha
yes right!!!
Nnesha
  • Nnesha
\[(1+\sin(x))^2 = (1+\sin)(1+\sin)\] (1+sin)^2 is same as (1+sin)(1+sin) foil it
anonymous
  • anonymous
and what happens when i foil it?
Nnesha
  • Nnesha
you will get 3 terms :3
Nnesha
  • Nnesha
then you will be able to cancel out some stuff at the numerator :=)
anonymous
  • anonymous
so 1+2sinx+sin^2x ???
Nnesha
  • Nnesha
yes right \[\huge\rm \frac{ \color{ReD}{cos^2x}+1+2sinx+\sin^2x }{ cosx(1+sinx) }\] cos^2x= what ? familiar with the identity ?
Nnesha
  • Nnesha
\[\huge\rm sin^2 \theta+\cos^2\theta=1\] this one solve for cos^2
anonymous
  • anonymous
cos^2=1-sin^2 ?
Nnesha
  • Nnesha
replace cos^2 with 1-sin^2 and then simplify !
Nnesha
  • Nnesha
\[\huge\rm \frac{ \color{ReD}{1-sin^2x}+1+2sinx+\sin^2x }{ cosx(1+sinx) }\]
anonymous
  • anonymous
2+2sinx ?
Nnesha
  • Nnesha
yes! what is common factor ?
Nnesha
  • Nnesha
\[\huge\rm \frac{ 2+2sinx }{ \cos(x)(1+sinx) }\] 2+sinx what is common in both terms?
anonymous
  • anonymous
sinx?
Nnesha
  • Nnesha
no first term is 2 and 2nd one is 2sinx so sinx isn't common
anonymous
  • anonymous
so 2
Nnesha
  • Nnesha
yes right take out the 2 what will you get ?
anonymous
  • anonymous
sinx
Nnesha
  • Nnesha
nah
Nnesha
  • Nnesha
when you take out 2 from 2 what will you have left wth?
anonymous
  • anonymous
0
Nnesha
  • Nnesha
haha well no in other words you have to divide both terms with the common factor so \[\huge\rm \frac{ 2}{ 2} +\frac{ 2sinx }{ 2 } =2(??+???)\]
anonymous
  • anonymous
1+sinx ?
Nnesha
  • Nnesha
yes! and we should keep the common factor outside the parentheses \[\huge\rm \frac{ 2(1+sinx) }{ cosx(1+sinx) }\]
Nnesha
  • Nnesha
simplify! that
anonymous
  • anonymous
2/cosx
Nnesha
  • Nnesha
yes right and last step ?
anonymous
  • anonymous
i dont know
Nnesha
  • Nnesha
thats fine! 2/cosx is same as \[\huge\rm 2 \times \frac{ 1 }{ cosx }\]
Nnesha
  • Nnesha
and 1/cosx= ??
anonymous
  • anonymous
secx
Nnesha
  • Nnesha
so 2/cosx= ? :=)
anonymous
  • anonymous
EXCITING
anonymous
  • anonymous
i honestly have no idea how to do anything with identities so thank you
Nnesha
  • Nnesha
ikr! trig is FUN(well i hate graph)
Nnesha
  • Nnesha
my pleasure! :=) welcome to openstudy!
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @rami2021 salam alakom \(\color{blue}{\text{End of Quote}}\) Walikum salam and yeah arabic is my fvt language but all i know is *kafya halik ?? ;~;
Nnesha
  • Nnesha
sorry about that alejand :3
anonymous
  • anonymous
no problem!
anonymous
  • anonymous
i'm about to post another question, i have one left

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