pottersheep
  • pottersheep
Please help me prove this trig identity.... (secx - cosecx)/(tanx - cotx) = (tanx - cotx) /(secx - cosecx)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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myininaya
  • myininaya
i would write both sides in terms of sin(x) and cos(x) and clear the compound fractions
myininaya
  • myininaya
and then you might have to use some other identities
anonymous
  • anonymous
break up the LHS into \[\frac{sec x}{tan x - cot x} + \frac{-csc x}{tan x - cot x} \]

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pottersheep
  • 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
  • anonymous
my advice would be to multiply by one using \[\frac{sinx}{sinx} \]
anonymous
  • anonymous
he means turn sec x into \[\frac{1}{cos x} \] etc...
anonymous
  • anonymous
for multiplying by one like i said above, dont use sin...use something that will reduce everything...
pottersheep
  • pottersheep
hmmm
pottersheep
  • pottersheep
i tried turning tan into sin/cos and cot into cis/sin and then I got all tans and cos's too
anonymous
  • 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
  • Mertsj
|dw:1322797409581:dw|
myininaya
  • 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
  • Mertsj
Now. Use the same technique on the right side.
myininaya
  • myininaya
this doesn't look like an identity
anonymous
  • anonymous
oh my apologies
pottersheep
  • pottersheep
Wow thanks for all that help...its supposed to be an identity though? sigh....perhaps its teachers mistake
myininaya
  • myininaya
it is not
pottersheep
  • pottersheep
oh wait a minute if I change both sides then I can get them to be equal
myininaya
  • myininaya
try x=pi/3 both sides have different outputs
myininaya
  • myininaya
one side is approximately 1.366 the other side is .73205
Mertsj
  • 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
  • myininaya
the equation does hold for some values but not all which makes it not an identity
pottersheep
  • pottersheep
Thanks you guys =) i guess i need much more practice
anonymous
  • 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
  • anonymous
start with|dw:1322798230411:dw|
Mertsj
  • Mertsj
So how did you get the denominator to be a product instead of a difference?
myininaya
  • 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
  • myininaya
so you wouldn't have that product anymore
pottersheep
  • pottersheep
yeah
pottersheep
  • 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
  • pottersheep
i still have like 18 questions to go...sigh...
pottersheep
  • pottersheep
trig identities :P not functions
myininaya
  • 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
  • Mertsj
Also, potter, are you absolutely sure that you copied the problem correctly?
pottersheep
  • pottersheep
Nope I understand...thank you everyone :) now i need to practice it
myininaya
  • myininaya
ok have fun
pottersheep
  • pottersheep
lol you guys are calling me potter hehe. and yeah but our teacher does have a bad reputation of making typos lol
myininaya
  • myininaya
is potter okay?
pottersheep
  • pottersheep
yep potter's fine :)
myininaya
  • myininaya
harry potter? lol is that what you were thinking?
pottersheep
  • pottersheep
yeah xD
myininaya
  • myininaya
it wouldn't be so bad to be the most powerful wizard of all time
pottersheep
  • pottersheep
yeah :( then i woulnt have to worry about these math problems lol
myininaya
  • myininaya
you could just turn a math problem you didn't feel like working on into a bunny
pottersheep
  • pottersheep
LOL hahaha
pottersheep
  • pottersheep
too bad i cant give u another medal for that :P

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