Question

Prove: (secx+cscx)cotx=1

Answer 1

So what do you think you should do first?

Answer 2

I did convert secx into 1/cosx and cscx into 1/sinx

Answer 3

and cotx into cosx/sinx

Answer 4

good start. So we want to get rid of those parentheses now...

Answer 5

it's going to turn into 1/sinx + cosx/sin^2x
I have no idea how it's going to get to 1

Answer 6

trust me?

Answer 7

Once you do that, get a common denominator

Answer 8

I dont know what the next step should be...

Answer 9

cosx/sinx+sin^2x

Answer 10

wait, that fraction isn't right look at it and use this \frac{}{}\) put your numerator and denominator in the brackets and just add a \( in front

Answer 11

actually it turns to (1+cosx)/(sinx+sin^2x) opps..

Answer 12

no... Idk how you got that. The common denom is just \(sin^2x\)

Answer 13

in the beginning cotx secx + cotx cscx =(cosx/sinx)*(1/cosx) + (cosx/sinx) (1/sinx)=
1/sinx + (cosx/sin^2x)

Answer 14

yea, now get a common denominator

Answer 15

that'll simplify to 1+cosx/sinx+sin^2x right?

Answer 16

no

Answer 17

How are you doing that?

Answer 18

1/sinx + (cosx/sin^2x)= (1+cosx)/(sinx+sin^2x)

Answer 19

no not at all ok I see what you did now

Answer 20

so rules for fractions

Answer 21

let a,b,c,d be numbers if we have \[\frac{a}{b}+\frac{c}{d}\not= \frac{a+b}{c+d}\]

Answer 22

I did a + c/b + d

Answer 23

Still not how we do it. We actually have to find the same denominator so we do this: \[\frac{a}{b}+\frac{c}{d}=(\frac{a}{b} \times \frac{d}{d}) +(\frac{c}{d} \times \frac{b}{b})= \frac{ad+cb}{bd}\]

Answer 24

this is called finding the least common denominator. Same as if you had 1/3+1/4

Answer 25

But yea, oops alphabetical moment on the wrong way

Answer 26

so it's (sin^2x+cosx sinx)/ sin^3x

Answer 27

no, well yes, but reduce

Answer 28

I have no clue what to do next then....

Answer 29

reduce

Answer 30

how?..

Answer 31

what term is common to each element?

Answer 32

said that backwards: Which element is common to every term?

Answer 33

sinx

Answer 34

good so eliminate or in dr who terms EXTERMINATE!!!

Answer 35

so it's sinx(sinx+cosx/sin^2x)

Answer 36

yup now here is where we find magic

Answer 37

I have no clue what the "magic" is now...

Answer 38

wait no read that wrong

Answer 39

there shouldnt be a sinx in front

Answer 40

they cancel with the bottom and =1

Answer 41

wait what do you mean?

Answer 42

so x/x=1 y/y=1 sinx/sinx=1

Answer 43

follow?

Answer 44

yea...

Answer 45

so our fraction becomes \[\frac{sinx(sinx+cosx)}{sinx(sin^2x)}\]

Answer 46

cancels and voila just \frac{(sinx+cosx)}{(sin^2x)}

Answer 47

\[\frac{(sinx+cosx)}{(sin^2x)}\]

Answer 48

yea...then what's next...

Answer 49

do you see the math there?

Answer 50

yep

Answer 51

like that we just did nothing new

Answer 52

k

Answer 53

meh h/o let me actually do this real quick It didn't simplify how I wanted it to in my head

Answer 54

are you sure you copied the problem correctly?

Answer 55

can you double check for me please?

Answer 56

I think the sec and csc might have been squared

Answer 57

nope they werent

Answer 58

hmmm...

Answer 59

tricky tricky

Answer 60

This is not true

Answer 61

teacher said it was a level 1 problem; I don't believe him...

Answer 62

This actually isn't plausible you effectively say that : \[\frac{(sinx+cosx)}{(sin^2x)}=1\]

Answer 63

which cannot be true

Answer 64

for every x

Answer 65

only if sinx=1 does this work

Answer 66

so hmm...

Answer 67

they said it was solvable?

Answer 68

yea, I was supposed to prove it but, it's complicated..

Answer 69

it's impossible to prove, just test it from the get go. use any x other than pi/2. it will not =1

Answer 70

i mean we could try some tricks... let's see

Answer 71

\[ \frac{\sqrt{sin^2x+cos^2x+2sinx cosx}}{sin^x}\]

Answer 72

oops\[\frac{\sqrt{sin^2x+cos^2x+2sinx cosx}}{sin^2x}\]

Answer 73

\[=\frac{\sqrt{2sinxcosx}}{sin^2x}\]

Answer 74

\[==\sqrt{\frac{2sinxcosx}}{sin^4x}}

Answer 75

\[=\sqrt{\frac{2sinxcosx}{sin^4x}}\]

Answer 76

\[=\sqrt{\frac{2cosx}{sin^2xsinx}}\]

Answer 77

\[=sinx\sqrt{2cotx}\]
Which obviously doesn not =1

Answer 78

my head is about to explode, it's 11pm and that's only problem #2 and I have 18 more questions to go.

Answer 79

I've tried starting with sinx/cosx(1/sinx+cosx)

Answer 80

It's physically impossible

Answer 81

just provide a counter ex from the start. Use like pi/4 and show it doesnt work

Answer 82

I mean that should show it. \[\frac{2}{\sqrt{2}}\times 1 \not= 1\]

Answer 83

and tell them they screwed up

Answer 84

go to the next question and realize your teacher is a jerk or a nitwit. And finish them. Don't sweat this one. Prove it wrong then move on

Answer 85

lol, my teacher's already a jerk

Answer 86

yea, when they said this was a level 1 proof, they must have meant a level one to disprove

Answer 87

probably just did it to give themselves an ego boost

Answer 88

if I have a sinx/sin^2x does it become 1/sinx?

Answer 89

yes,but that still isnt 1

Answer 90

My other problems is 1/1-sinx = 1 +sinx/ cos^2x

Answer 91

that I can buy

Answer 92

so 1 +sinx/ cos^2x will be 1+sinx/1-sin^2x, I need to get rid of the sinx/sinx

Answer 93

do you mind posting it as another q though so I can get another medal?

Answer 94

okay