Question

prove
please help
cscx-sinx=cotxcosx

Answer 1

Start by breaking them down into sin and cos. Should make it easier to solve.

Answer 2

i got to (1/sinx)-(1/cscx)
i dont know what to do from there

Answer 3

i got to (1cscx-1sinx)/(sinxcscx)

Answer 4

Of course I say that and now I can't solve it. One sec.

Answer 5

k

Answer 6

leave one sinx as it is.
Try to simplify by writing it as
(1/sinx) - sinx

Answer 7

but i don't know what to do from there :O

Answer 8

oh... well
\[\Large \sin(x) = \frac{\sin^2(x)}{\sin(x)}\]
did that only for the sake of giving them the same denominator

Answer 9

<whistles...>
\[\Large \frac1{\sin(x) }- \frac{\sin^2(x)}{\sin(x)}=\color{red}?\]
HINT: They have the same denominator now, you may simply subtract the numerators...

Answer 10

is there another way of doing it?

Answer 11

None that I'm familiar with... consult with our local magician :3
@shubhamsrg ?

Answer 12

you may convert RHS into LHS :|

Answer 13

Wow. Thank you terenzreignz for coming along.

Answer 14

i'm doing this online on a program

Answer 15

You can call me Terence... or TJ if it's too much of a bother... and no worries, @amishjeb you laid the foundation to this question... so to speak :)

Answer 16

Terenzreigz has the LHS solved and you'll need to get the RHS looking the same. If you go back to putting things into sin and cos you can get there easy. You'll also need to use a the basic pythagorean theorem, but solve for cos^2.

Answer 17

(1/sinx )-sinx

it won't me let it type the way you told me

Answer 18

@amishjeb It's far more 'magical' to just leave one side alone and turn the other side into that...
\[\Large \frac1{\sin(x) }- \frac{\sin^2(x)}{\sin(x)}=\color{red}?\]
So... we can combine them into one fraction...
\[\Large \frac{1-\sin^2(x)}{\sin(x) }\]

And @angelita321 ... does THIS
\[\Large \frac{\color{blue}{1-\sin^2(x)}}{\sin(x) }\]
Look familiar to you?

Answer 19

where do you get the sin^(2)x/sinx from???

Answer 20

Well, granted that
\[\Large 1 \cdot \sin(x) = \sin(x)\]
I just uhh... thought about the fact that \(\Large \frac{\sin(x)}{\sin(x)}=1\)
So that we get
\[\Large \color{red}1\cdot \sin(x) =\color{red}{\frac{\sin(x)}{\sin(x)}}\cdot \sin(x) =\frac{\sin^2(x)}{\sin(x)}\]

Answer 21

Problem? ^_^

Answer 22

than you get cos^(2)x/sinx ???

Answer 23

Very good :P
\[\Large \frac{\cos^2(x)}{\sin(x)}\]

Answer 24

than?

Answer 25

And therefore...?
Well, we can always split the \(\cos^2(x)\)\[\Large \frac{\cos(x)\cos(x)}{\sin(x)}\]
Take it away , @angelita321

Answer 26

cosxcotx

Answer 27

And thus, it ends :P
Wasn't that just 'magical' ? XD

Answer 28

yes!!l
this problems are confusing >.< and i still need to do 2 more (T-T)

Answer 29

Post the new ones, quickly.
We have no idea when OS gets down :P

Answer 30

cscx-cotxcosx=sinx

Answer 31

\[(1/sinx) -sinx=(1-\sin^2x)/sinx=\cos^2x/sinx=cotxcosx\]

Answer 32

Once again, reduce everything (on the left-side) to sines and cosines...

Answer 33

1/sinx - cotxcosx

Answer 34

do something about that cot x

Answer 35

change it to 1/tanx ?

Answer 36

better for it to be in terms of sin(x) and cos(x)

Answer 37

Stuck? :)
\[\Large \cot(x) = \frac{\cos(x)}{\sin(x)}\]
work from there :P

Answer 38

cosx crosses out

Answer 39

cos^(2)x / sinx

Answer 40

There you go... so on the left, we have
\[\Large \frac1{\sin(x)}-\frac{\cos^2(x)}{\sin(x)}\]
Once again, they have the same denominator, the numerators may simply be subtracted, leaving you with...?

Answer 41

1/sinx - 1-sinx/sinx

Answer 42

huh?
How did you arrive at that?

Answer 43

i don't know

Answer 44

Take it slow...
We have this\[\Large \frac1{\sin(x)}-\frac{\cos^2(x)}{\sin(x)}\]right?

Combine them into one fraction, as you would for fractions with the same denominator...
\[\large \frac{a}c-\frac{b}c=\frac{a-b}c\]

Answer 45

1-cos^(2)x / sinx

Answer 46

mhmm
once again, THIS
\[\Large \frac{\color{blue}{1-\cos^2(x)}}{\sin(x)}\]
should look familiar...

Answer 47

it doesn't let me

Answer 48

doesn't matter, just type as you normally do, I'll understand :)

Answer 49

holdon on, let me post the pic

Answer 50

this show up

Answer 51

can't read it

Answer 52

What does it say?

Answer 53

i fixed it so it can be 1-cos^(2)x/sinx but from there i tried putting sin^(2)x/sinx

Answer 54

Well, that's not it, isn't it?
\[\Large \frac{\sin^2(x)}{\sin(x)}=\frac{\cancel{\sin(x)}\sin(x)}{\cancel{\sin(x)}}=\sin(x)\]

Answer 55

yayy it said it was correct :D

Answer 56

one more to go!!
secx-sin^(2)xsecx=cosx

Answer 57

1/cosx-sin^(2)xsecx

Answer 58

Do something about that sec(x)

Answer 59

both?

Answer 60

On second thought, return it to the original...
\[\Large \sec(x)-\sin^2(x) \sec(x) \]and just factor out the sec(x)

Answer 61

You get...?

Answer 62

secx(1-sin^(2)x)

Answer 63

later on do you get secx(cos^(2)x)

Answer 64

Yes you do...
now, that is just equal to
\[\Large \color{green}{\sec(x)\cos(x)}\cos(x)\]And that part in green is just equal to...?

Answer 65

????

Answer 66

Don't ???? me :P
What is sec(x) in terms of sines and cosines? :D

Answer 67

(1/cosx)cosxcosx

crosses out
cosx ??

Answer 68

Yes... one of the cos(x) crosses out... leaving...?
:P

Answer 69

i don't like proving identity. more of this problems pop out in the webpage im doing >.<

Answer 70

what do we do if the problems are bigger

(sinx)/(1+cosx) + (1+cosx)/(sinx) = 2cscx

Answer 71

How many of these do you have? -.-
And anyway, if you have many fractions, try to combine them into one...

Answer 72

i'm not sure. it's part of a program i'm in and i have to do at least 80% of pre-cal/trig. and random things pop out. i'm in 77% right now v.v so i have to keep doing this (T-T)

Answer 73

reciprocal of sin =1/cscx for both