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anonymous
 4 years ago
6) If sec theta+tan theta = P. PT sin theta=P^21/P^2+1
anonymous
 4 years ago
6) If sec theta+tan theta = P. PT sin theta=P^21/P^2+1

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

myininaya
 4 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 \[PT \sin(\theta)=\frac{P^21}{P^2+1}\]?

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

myininaya
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.1i havent solved it yet

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

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

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

myininaya
 4 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
 4 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
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.1\[\sin(\theta)=\frac{P^21}{P^2+1}\]

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

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

anonymous
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.1hey zarkon i had to think on it isn't that crazy?

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

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

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

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

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

myininaya
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.1zarkon i did this for you :)

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

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

anonymous
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.2\[1+\sin(\theta)=p\cos(\theta)\]

myininaya
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.2\[1+\sin(\theta)=p\sqrt{1\sin^2(\theta)}\]

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

Zarkon
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.1you going to write is as a quadratic

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

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

Zarkon
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.2\[1p^2+2x+(1+p^2)x^2=0\]

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

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

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

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