abichu
6) If sec theta+tan theta = P. PT sin theta=P^2-1/P^2+1
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myininaya
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we want to show the second part?
abichu
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yea!
myininaya
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what is T?
abichu
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PT= PROVE THAT
myininaya
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so we get to suppose
\[\sec(\theta)+\tan(\theta)=P\]
and we want to show
\[PT \sin(\theta)=\frac{P^2-1}{P^2+1}\]?
myininaya
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oh ok
abichu
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and guys can you answer my prev.Questions too?! please..
myininaya
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\[\sec(\theta)+\tan(\theta)=\frac{1}{\cos(\theta)}+\frac{\sin(\theta)}{\cos(\theta)}=\frac{1+\sin(\theta)}{\cos(\theta)}=P\]
thinking....
myininaya
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\[\frac{1+\sin(\theta)}{\cos(\theta)}*\frac{1-\sin(\theta)}{1-\sin(\theta)}=\frac{1-\sin^2(\theta)}{\cos(\theta)(1-\sin(\theta))}\]
\[=\frac{\cos^2(\theta)}{\cos(\theta)(1-\sin(\theta))}=\frac{\cos(\theta)}{1-\sin(\theta)}=P\]
hmmm....
abichu
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solved?
myininaya
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no lol
myininaya
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i havent solved it yet
myininaya
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maybe i should try this on paper first
abichu
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what bout my prev.sums? can you solve it?
myininaya
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let me try this one first
abichu
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sure! :)
myininaya
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if \[\sec(\theta)+\tan(\theta)=P\] then \[(\sec(\theta)+\tan(\theta))(\sec(\theta)-\tan(\theta))=P(\sec(\theta)-\tan(\theta))\]
but \[(\sec(\theta)+\tan(\theta))(\sec(\theta)-\tan(\theta))=\sec^2(\theta)-\tan^2(\theta)=1\]
so we have
\[1=P(\sec(\theta)-\tan(\theta))\]
=>
\[\sec(\theta)-\tan(\theta)=\frac{1}{P}\]
lets see if we can use this
myininaya
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lets see
we have
that
\[\sec(\theta)+\tan(\theta)=\frac{1+\sin(\theta)}{\cos(\theta)}\]
and
\[\sec(\theta)-\tan(\theta)=\frac{1-\sin(\theta)}{\cos(\theta)}\]
adding these together we get
\[\frac{1+\sin(\theta)}{\cos(\theta)}+\frac{1-\sin(\theta)}{\cos(\theta)}=P+\frac{1}{P}\]
\[\frac{2}{\cos(\theta)}=\frac{P^2+1}{P}\]
\[\frac{2P}{P^2+1}=\cos(\theta)\]
myininaya
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but if we imagine we have a right triangle
and we draw theta somewhere
then cos tells adj/hyp=2p/(P^2+1)
we can find the opposite of side of theta
opp^2+adj^2=hyp^2
\[opp^2=hyp^2-adj^2=(P^2+1)^2-(2P)^2=P^4+2P^2+1-2^2P^2\]
so we have
\[opp=\sqrt{P^4-2P^2+1}\]
but
sin=opp/hyp
myininaya
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\[\sin(\theta)=\frac{\sqrt{P^4-2P^2+1}}{P^2+1}=\frac{\sqrt{(P^2-1)^2}}{P^2+1}\]
myininaya
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\[\sin(\theta)=\frac{P^2-1}{P^2+1}\]
myininaya
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any questions?
abichu
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yea..can you solve the other problems?
myininaya
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zarkon?
abichu
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what?
myininaya
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i was looking for someone
Zarkon
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yes
abichu
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and can you temme one thing..can you temme the stepin correct order so that i can copy it in my rough boook! ;)
myininaya
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hey zarkon
i had to think on it
isn't that crazy?
Zarkon
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yes
myininaya
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the problem
Zarkon
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trying to think if there is an easier way
myininaya
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i have no clue it was hard for me to see that way
myininaya
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what is temme?
myininaya
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i don't know that word
myininaya
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use small words please lol
myininaya
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well i mean i guess its small but i don't know it
myininaya
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tell me?
myininaya
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temme=tell me?
myininaya
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i thought i showed u the steps
zarkon thinks there may be an easier way though
myininaya
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zarkon i did this for you
:)
myininaya
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lol
Zarkon
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;)
Zarkon
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how about this...
myininaya
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and for abichu
dont feel left out
Zarkon
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get to \[\frac{1+sin(\theta)}{\cos(\theta)}=p\]
...
abichu
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i am soo weak in math..that's why am seeking help from you guys! if you dont mind help me! ;)
myininaya
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we are helping
Zarkon
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\[1+\sin(\theta)=p\cos(\theta)\]
myininaya
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i showed you one way
and zarkon is gonna show you another it appears
Zarkon
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\[1+\sin(\theta)=p\sqrt{1-\sin^2(\theta)}\]
Zarkon
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square both sides...
myininaya
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omg i see where you are going
Zarkon
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\[1+2\sin(\theta)+\sin^2(\theta)=p^2(1-\sin^2(\theta))\]
myininaya
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you going to write is as a quadratic
myininaya
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and solve for sin(theta)
abichu
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can you tel me from the 1st step if you dot mind?!
Zarkon
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move stuff to one side
\[1-p^2+2\sin(\theta)+(1+p^2)\sin^2(\theta)=0\]
myininaya
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brillant!
Zarkon
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\[1-p^2+2x+(1+p^2)x^2=0\]
Zarkon
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solve this quadratic ;)
Zarkon
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fun stuff :)
myininaya
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yes
myininaya
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you are a showoff lol
Zarkon
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lol...now abichu has two ways to do it
abichu
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i donno how to solve it quadratic..can you doo it n give?
myininaya
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\[x=\frac{-2 \pm \sqrt{(2)^2-4(1+p^2)(1-p^2)}}{2(1+p^2)}\]