Please help me prove this trig identity.... (secx - cosecx)/(tanx - cotx) = (tanx - cotx) /(secx - cosecx)

- pottersheep

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- chestercat

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- myininaya

i would write both sides in terms of sin(x) and cos(x) and clear the compound fractions

- myininaya

and then you might have to use some other identities

- anonymous

break up the LHS into \[\frac{sec x}{tan x - cot x} + \frac{-csc x}{tan x - cot x} \]

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## More answers

- pottersheep

Oh....I'll try it that way then but Im not sure if I can reach the answer still.... And how can I change them to cos and sin? they are not squared :(

- anonymous

my advice would be to multiply by one using \[\frac{sinx}{sinx} \]

- anonymous

he means turn sec x into \[\frac{1}{cos x} \] etc...

- anonymous

for multiplying by one like i said above, dont use sin...use something that will reduce everything...

- pottersheep

hmmm

- pottersheep

i tried turning tan into sin/cos and cot into cis/sin and then I got all tans and cos's too

- anonymous

try taking everything in terms of cos and sin and then simplifying as much as you can. Don't leave any sec, csc, tan or cot

- Mertsj

|dw:1322797409581:dw|

- myininaya

i'm a her \[\frac{\frac{1}{\cos(x)}-\frac{1}{\sin(x)}}{\frac{\sin(x)}{\cos(x)}-\frac{\cos(x)}{\sin(x)}}=\frac{\frac{\sin(x)}{\cos(x)}-\frac{\cos(x)}{\sin(x)}}{\frac{1}{\cos(x)}-\frac{1}{\sin(x)}}\] now for both sides we want to clear the compound fraction action going on \[\frac{\frac{1}{\cos(x)}-\frac{1}{\sin(x)}}{\frac{\sin(x)}{\cos(x)}-\frac{\cos(x)}{\sin(x)}} \cdot \frac{\cos(x)\sin(x)}{\cos(x)\sin(x)}=\frac{\frac{\sin(x)}{\cos(x)}-\frac{\cos(x)}{\sin(x)}}{\frac{1}{\cos(x)}-\frac{1}{\sin(x)}} \cdot \frac{\cos(x)\sin(x)}{\cos(x)\sin(x)} \] \[\frac{\sin(x)-\cos(x)}{\sin^2(x)-\cos^2(x)}=\frac{\sin^2(x)-\cos^2(x)}{\sin(x)-\cos(x)}\] now we can some factoring on both sides \[\frac{\sin(x)-\cos(x)}{(\sin(x)-\cos(x))(\sin(x)+\cos(x))}=\frac{(\sin(x)+\cos(x))(\sin(x)-\cos(x))}{\sin(x)-\cos(x)}\] so we have \[\frac{1}{\sin(x)+\cos(x)}=\sin(x)+\cos(x)\] ...

- Mertsj

Now. Use the same technique on the right side.

- myininaya

this doesn't look like an identity

- anonymous

oh my apologies

- pottersheep

Wow thanks for all that help...its supposed to be an identity though? sigh....perhaps its teachers mistake

- myininaya

it is not

- pottersheep

oh wait a minute if I change both sides then I can get them to be equal

- myininaya

try x=pi/3 both sides have different outputs

- myininaya

one side is approximately 1.366 the other side is .73205

- Mertsj

No. It is not. You have shown that they indeed are reciprocals which was apparent by a casual inspection of the original problem.

- myininaya

the equation does hold for some values but not all which makes it not an identity

- pottersheep

Thanks you guys =) i guess i need much more practice

- anonymous

when I simplified both sides after making them in terms of sin and cos I got (sinx-cosx)(sinx^2-cosx^2) ------------------------ cosx^2*sinx^2 for both the left hand side and the right hand side

- anonymous

start with|dw:1322798230411:dw|

- Mertsj

So how did you get the denominator to be a product instead of a difference?

- myininaya

maybe he combine the fractions on top and on bottom but cos(x)sin(x) in the denominator of the bottom fraction would cancel with cos(x)sin(x) with the top's denominator

- myininaya

so you wouldn't have that product anymore

- pottersheep

yeah

- pottersheep

omg! this is the second day we started trig functions and our teacher barely taught us anything......i wish she would start off with simpler question....

- pottersheep

i still have like 18 questions to go...sigh...

- pottersheep

trig identities :P not functions

- myininaya

anyways potter we have already shown that this is not an identity do you have any questions as to why we concluded that?

- Mertsj

Also, potter, are you absolutely sure that you copied the problem correctly?

- pottersheep

Nope I understand...thank you everyone :) now i need to practice it

- myininaya

ok have fun

- pottersheep

lol you guys are calling me potter hehe. and yeah but our teacher does have a bad reputation of making typos lol

- myininaya

is potter okay?

- pottersheep

yep potter's fine :)

- myininaya

harry potter? lol is that what you were thinking?

- pottersheep

yeah xD

- myininaya

it wouldn't be so bad to be the most powerful wizard of all time

- pottersheep

yeah :( then i woulnt have to worry about these math problems lol

- myininaya

you could just turn a math problem you didn't feel like working on into a bunny

- pottersheep

LOL hahaha

- pottersheep

too bad i cant give u another medal for that :P

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