Question

6) If sec theta+tan theta = P. PT sin theta=P^2-1/P^2+1

Answer 1

we want to show the second part?

Answer 2

yea!

Answer 3

what is T?

Answer 4

PT= PROVE THAT

Answer 5

so we get to suppose
\[\sec(\theta)+\tan(\theta)=P\]
and we want to show
\[PT \sin(\theta)=\frac{P^2-1}{P^2+1}\]?

Answer 6

oh ok

Answer 7

and guys can you answer my prev.Questions too?! please..

Answer 8

\[\sec(\theta)+\tan(\theta)=\frac{1}{\cos(\theta)}+\frac{\sin(\theta)}{\cos(\theta)}=\frac{1+\sin(\theta)}{\cos(\theta)}=P\]
thinking....

Answer 9

\[\frac{1+\sin(\theta)}{\cos(\theta)}*\frac{1-\sin(\theta)}{1-\sin(\theta)}=\frac{1-\sin^2(\theta)}{\cos(\theta)(1-\sin(\theta))}\]
\[=\frac{\cos^2(\theta)}{\cos(\theta)(1-\sin(\theta))}=\frac{\cos(\theta)}{1-\sin(\theta)}=P\]
hmmm....

Answer 10

solved?

Answer 11

no lol

Answer 12

i havent solved it yet

Answer 13

maybe i should try this on paper first

Answer 14

what bout my prev.sums? can you solve it?

Answer 15

let me try this one first

Answer 16

sure! :)

Answer 17

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

Answer 18

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
\[\frac{1+\sin(\theta)}{\cos(\theta)}+\frac{1-\sin(\theta)}{\cos(\theta)}=P+\frac{1}{P}\]
\[\frac{2}{\cos(\theta)}=\frac{P^2+1}{P}\]
\[\frac{2P}{P^2+1}=\cos(\theta)\]

Answer 19

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

Answer 20

\[\sin(\theta)=\frac{\sqrt{P^4-2P^2+1}}{P^2+1}=\frac{\sqrt{(P^2-1)^2}}{P^2+1}\]

Answer 21

\[\sin(\theta)=\frac{P^2-1}{P^2+1}\]

Answer 22

any questions?

Answer 23

yea..can you solve the other problems?

Answer 24

zarkon?

Answer 25

what?

Answer 26

i was looking for someone

Answer 27

yes

Answer 28

and can you temme one thing..can you temme the stepin correct order so that i can copy it in my rough boook! ;)

Answer 29

hey zarkon
i had to think on it
isn't that crazy?

Answer 30

yes

Answer 31

the problem

Answer 32

trying to think if there is an easier way

Answer 33

i have no clue it was hard for me to see that way

Answer 34

what is temme?

Answer 35

i don't know that word

Answer 36

use small words please lol

Answer 37

well i mean i guess its small but i don't know it

Answer 38

tell me?

Answer 39

temme=tell me?

Answer 40

i thought i showed u the steps
zarkon thinks there may be an easier way though

Answer 41

zarkon i did this for you
:)

Answer 42

lol

Answer 43

;)

Answer 44

how about this...

Answer 45

and for abichu
dont feel left out

Answer 46

get to \[\frac{1+sin(\theta)}{\cos(\theta)}=p\]
...

Answer 47

i am soo weak in math..that's why am seeking help from you guys! if you dont mind help me! ;)

Answer 48

we are helping

Answer 49

\[1+\sin(\theta)=p\cos(\theta)\]

Answer 50

i showed you one way
and zarkon is gonna show you another it appears

Answer 51

\[1+\sin(\theta)=p\sqrt{1-\sin^2(\theta)}\]

Answer 52

square both sides...

Answer 53

omg i see where you are going

Answer 54

\[1+2\sin(\theta)+\sin^2(\theta)=p^2(1-\sin^2(\theta))\]

Answer 55

you going to write is as a quadratic

Answer 56

and solve for sin(theta)

Answer 57

can you tel me from the 1st step if you dot mind?!

Answer 58

move stuff to one side

\[1-p^2+2\sin(\theta)+(1+p^2)\sin^2(\theta)=0\]

Answer 59

brillant!

Answer 60

\[1-p^2+2x+(1+p^2)x^2=0\]

Answer 61

solve this quadratic ;)

Answer 62

fun stuff :)

Answer 63

yes

Answer 64

you are a showoff lol

Answer 65

lol...now abichu has two ways to do it

Answer 66

i donno how to solve it quadratic..can you doo it n give?

Answer 67

\[x=\frac{-2 \pm \sqrt{(2)^2-4(1+p^2)(1-p^2)}}{2(1+p^2)}\]