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myininaya Group TitleBest ResponseYou've already chosen the best response.1
we want to show the second part?
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
what is T?
 3 years ago

abichu Group TitleBest ResponseYou've already chosen the best response.0
PT= PROVE THAT
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
so we get to suppose \[\sec(\theta)+\tan(\theta)=P\] and we want to show \[PT \sin(\theta)=\frac{P^21}{P^2+1}\]?
 3 years ago

abichu Group TitleBest ResponseYou've already chosen the best response.0
and guys can you answer my prev.Questions too?! please..
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
\[\sec(\theta)+\tan(\theta)=\frac{1}{\cos(\theta)}+\frac{\sin(\theta)}{\cos(\theta)}=\frac{1+\sin(\theta)}{\cos(\theta)}=P\] thinking....
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{1+\sin(\theta)}{\cos(\theta)}*\frac{1\sin(\theta)}{1\sin(\theta)}=\frac{1\sin^2(\theta)}{\cos(\theta)(1\sin(\theta))}\] \[=\frac{\cos^2(\theta)}{\cos(\theta)(1\sin(\theta))}=\frac{\cos(\theta)}{1\sin(\theta)}=P\] hmmm....
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
i havent solved it yet
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
maybe i should try this on paper first
 3 years ago

abichu Group TitleBest ResponseYou've already chosen the best response.0
what bout my prev.sums? can you solve it?
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
let me try this one first
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
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
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
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 \[\frac{1+\sin(\theta)}{\cos(\theta)}+\frac{1\sin(\theta)}{\cos(\theta)}=P+\frac{1}{P}\] \[\frac{2}{\cos(\theta)}=\frac{P^2+1}{P}\] \[\frac{2P}{P^2+1}=\cos(\theta)\]
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
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 \[opp^2=hyp^2adj^2=(P^2+1)^2(2P)^2=P^4+2P^2+12^2P^2\] so we have \[opp=\sqrt{P^42P^2+1}\] but sin=opp/hyp
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
\[\sin(\theta)=\frac{\sqrt{P^42P^2+1}}{P^2+1}=\frac{\sqrt{(P^21)^2}}{P^2+1}\]
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
\[\sin(\theta)=\frac{P^21}{P^2+1}\]
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
any questions?
 3 years ago

abichu Group TitleBest ResponseYou've already chosen the best response.0
yea..can you solve the other problems?
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
i was looking for someone
 3 years ago

abichu Group TitleBest ResponseYou've already chosen the best response.0
and can you temme one thing..can you temme the stepin correct order so that i can copy it in my rough boook! ;)
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
hey zarkon i had to think on it isn't that crazy?
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
the problem
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
trying to think if there is an easier way
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
i have no clue it was hard for me to see that way
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
what is temme?
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
i don't know that word
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
use small words please lol
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
well i mean i guess its small but i don't know it
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
temme=tell me?
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
i thought i showed u the steps zarkon thinks there may be an easier way though
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
zarkon i did this for you :)
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
how about this...
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
and for abichu dont feel left out
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
get to \[\frac{1+sin(\theta)}{\cos(\theta)}=p\] ...
 3 years ago

abichu Group TitleBest ResponseYou've already chosen the best response.0
i am soo weak in math..that's why am seeking help from you guys! if you dont mind help me! ;)
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
we are helping
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
\[1+\sin(\theta)=p\cos(\theta)\]
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
i showed you one way and zarkon is gonna show you another it appears
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
\[1+\sin(\theta)=p\sqrt{1\sin^2(\theta)}\]
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
square both sides...
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
omg i see where you are going
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
\[1+2\sin(\theta)+\sin^2(\theta)=p^2(1\sin^2(\theta))\]
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
you going to write is as a quadratic
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
and solve for sin(theta)
 3 years ago

abichu Group TitleBest ResponseYou've already chosen the best response.0
can you tel me from the 1st step if you dot mind?!
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
move stuff to one side \[1p^2+2\sin(\theta)+(1+p^2)\sin^2(\theta)=0\]
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
brillant!
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
\[1p^2+2x+(1+p^2)x^2=0\]
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
solve this quadratic ;)
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
fun stuff :)
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
you are a showoff lol
 3 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
lol...now abichu has two ways to do it
 3 years ago

abichu Group TitleBest ResponseYou've already chosen the best response.0
i donno how to solve it quadratic..can you doo it n give?
 3 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
\[x=\frac{2 \pm \sqrt{(2)^24(1+p^2)(1p^2)}}{2(1+p^2)}\]
 3 years ago
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