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pottersheep
Please help me prove this trig identity.... (secx - cosecx)/(tanx - cotx) = (tanx - cotx) /(secx - cosecx)
i would write both sides in terms of sin(x) and cos(x) and clear the compound fractions
and then you might have to use some other identities
break up the LHS into \[\frac{sec x}{tan x - cot x} + \frac{-csc x}{tan x - cot x} \]
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 :(
my advice would be to multiply by one using \[\frac{sinx}{sinx} \]
he means turn sec x into \[\frac{1}{cos x} \] etc...
for multiplying by one like i said above, dont use sin...use something that will reduce everything...
i tried turning tan into sin/cos and cot into cis/sin and then I got all tans and cos's too
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
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 \[\frac{1}{\sin(x)+\cos(x)}=\sin(x)+\cos(x)\] ...
Now. Use the same technique on the right side.
this doesn't look like an identity
Wow thanks for all that help...its supposed to be an identity though? sigh....perhaps its teachers mistake
oh wait a minute if I change both sides then I can get them to be equal
try x=pi/3 both sides have different outputs
one side is approximately 1.366 the other side is .73205
No. It is not. You have shown that they indeed are reciprocals which was apparent by a casual inspection of the original problem.
the equation does hold for some values but not all which makes it not an identity
Thanks you guys =) i guess i need much more practice
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
start with|dw:1322798230411:dw|
So how did you get the denominator to be a product instead of a difference?
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
so you wouldn't have that product anymore
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....
i still have like 18 questions to go...sigh...
trig identities :P not functions
anyways potter we have already shown that this is not an identity do you have any questions as to why we concluded that?
Also, potter, are you absolutely sure that you copied the problem correctly?
Nope I understand...thank you everyone :) now i need to practice it
lol you guys are calling me potter hehe. and yeah but our teacher does have a bad reputation of making typos lol
yep potter's fine :)
harry potter? lol is that what you were thinking?
it wouldn't be so bad to be the most powerful wizard of all time
yeah :( then i woulnt have to worry about these math problems lol
you could just turn a math problem you didn't feel like working on into a bunny
too bad i cant give u another medal for that :P