Question

I'm a bit confused, I have no idea where to start ;-;

These are proofs btw. I need to prove if it is true/false

\[\csc^2x-1=\cot^2x\]

I think I need to start with changing csc to 1/y
Then change the 1 to \[\sin^2x + cos^2x\] because it equals one?

Some clarification would be nice, thanx c:

Answer 1

how would you write csc^2x in terms of sin , cos ??

Answer 2

\[\frac{ 1 }{ \sin }^2x\]

Answer 3

hmm the 2 is flying away :P

Answer 4

Haha, I'll rewrite it
\[(\frac{ 1 }{ \sin })^2x\]

Answer 5

\[\frac{ 1 }{ \sin^2x } -1\]
good
simplify
rewrite it as a single fraction ( find the common denominator )

Answer 6

That's how you rewrite it?

Answer 7

Ooooh, ok

Answer 8

I get it

Answer 9

So \[-\frac{ \sin }{ \sin }\]

Answer 10

yes
because of the fact csc^2x=1/sin^2x
i can replace csc^2x with 1/sin^2x

Answer 11

wait
\[\frac{ \sin^2x }{ \sin^2x }\]

Answer 12

hmm no lets say sinx= x
\[\frac{ 1 }{ x^2 }-1\] how would you solve this algebra expression ?? :)

Answer 13

cameron just a moment

do you know that cosec x = 1/ sin x
yes ?

Answer 14

Yes

Answer 15

so than cosec^2 x = (1/sin x)^2
yes ?

Answer 16

Yea

Answer 17

but (1/sin x)^2 = 1/sin^2 x

Answer 18

ok. ?

Answer 19

Yea, because when you square 1 it stays the same

Answer 20

exactly

Answer 21

Then you can turn 1 into sin^2x + cos^2x = 1

Answer 22

Cancel out the sines and you have cos

Answer 23

Right?

Answer 24

let me rewrite it \[\frac{ 1 }{ \sin^2x } -\frac{1}{1}= \cot^2x\]
we are working on the left side
to find the common denominator
multiply top and bottom of the fractions by sin^2x

Answer 25

sensitive i have one different idea without common denominator

Answer 26

rewrite the 1 from numerator in the form of sin^2 x +cos^2 x

Answer 27

\[\frac{ 1 }{ \sin^2x} *\frac{\sin^2x}{\sin^2x} - \frac{1}{1}*\frac{\sin^2x}{\sin^2x}=cot^2x\]

\[\frac{ 1 -\sin^2x}{ \sin^2x } =cot^2x\]
let me know if you have a question about this ??

Answer 28

yes thats the easy one

btw cameron there are multiple ways to prove trig function :)

Answer 29

gtg sorry

Answer 30

How did you get the multiplication of sin^2x/sin^2x?

Answer 31

Lol, thanks for helpng though :)

Answer 32

snsitive the first column of sin^2x/sin^2x not need you write there this is wrong

Answer 33

so bc. than will get the denominator sin^4 x

Answer 34

cameron do you see it about what i talk ?

Answer 35

I think I get it now
You start with changing csc^2x to 1/sin^2x
Then you change the -1 to sin^2x/sin^2x
Then You change the cot to cos^2x/sin^2x

Since everything is common denominator, you can simplify
Right? c:

Answer 36

no

the right part need remain so

and from the left part you need getting the right part

Answer 37

Change the 1 to sin^2x + cos^2x?

Answer 38

first the 1 rewrite in the form of sin^2 /sin^2 x
and after this rewrite the 1 from numerator of sin^2 x in this form

Answer 39

cameron do you understand it now ?

Answer 40

I got it!
\[\frac{ 1 }{ \sin^2x }-\frac{ \sin^2x }{ \sin^2x }=\cot^2x\]
\[\frac{ 1-\sin^2x }{ \sin^2x }\rightarrow \frac{ \sin^2x+\cos^2x-\sin^2x }{ \sin^2x }\rightarrow \frac{ \cos^2x }{ \sin^2x }=\cot^2x\]
Right? c:

Answer 41

yes this is the right easy way how i thought it

hope helped

Answer 42

Thank you so much <3
I didn't understand it, thank you for helping me so c;

Answer 43

was my pleasure

what you dont understand there ? - please ask me now

Answer 44

I get everything c:
Thank you for asking

Answer 45

ok. so good luck

bye bye

Answer 46

i wrote it correctly that's how she got the correct answer:) \[\frac{ 1 }{ \sin^2x } *\frac{\sin^2x}{\sin^2x} - \frac{1}{1} *\frac{\sin^2x}{\sin^2x}\]
\[\frac{ 1 }{ \cancel{\sin^2x} } *\frac{\cancel{\sin^2x}}{\sin^2x} - \frac{1}{1} *\frac{\sin^2x}{\sin^2x}\]
we will get \[\frac{ 1 }{ \sin^2x } -\frac{\sin^2x}{\sin^2x}\]
now you can rewrite as a single fraction\[\frac{ 1-\sin^2x }{ \sin^2x }\]
i like how you replaced 1 with sin^2x+cos^2x or
you can solve this equation for cos^2x
\[\sin^2x+\cos^2x=1 \]
\[\cos^2x=1-\sin^2x\]
replace the numerator by cos^2x
\[\frac{ 1-\sin^2x }{ \sin^2x }=\frac{ cos^2x }{ \sin^2x }\]
which is equal to cot^2x