A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

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

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

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

myininaya
 3 years ago
Best 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....

myininaya
 3 years ago
Best 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....

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

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

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

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

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1if \[\sec(\theta)+\tan(\theta)=P\] 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
 3 years ago
Best ResponseYou've already chosen the best response.1lets 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
 3 years ago
Best ResponseYou've already chosen the best response.1but 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1\[x=\frac{2 \pm \sqrt{(2)^24(1+p^2)(1p^2)}}{2(1+p^2)}\]
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.