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pottersheep
Group Title
Please help me prove this trig identity....
(secx  cosecx)/(tanx  cotx) = (tanx  cotx) /(secx  cosecx)
 2 years ago
 2 years ago
pottersheep Group Title
Please help me prove this trig identity.... (secx  cosecx)/(tanx  cotx) = (tanx  cotx) /(secx  cosecx)
 2 years ago
 2 years ago

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myininaya Group TitleBest ResponseYou've already chosen the best response.3
i would write both sides in terms of sin(x) and cos(x) and clear the compound fractions
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
and then you might have to use some other identities
 2 years ago

moses_ Group TitleBest ResponseYou've already chosen the best response.0
break up the LHS into \[\frac{sec x}{tan x  cot x} + \frac{csc x}{tan x  cot x} \]
 2 years ago

pottersheep Group TitleBest ResponseYou've already chosen the best response.1
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 :(
 2 years ago

moses_ Group TitleBest ResponseYou've already chosen the best response.0
my advice would be to multiply by one using \[\frac{sinx}{sinx} \]
 2 years ago

moses_ Group TitleBest ResponseYou've already chosen the best response.0
he means turn sec x into \[\frac{1}{cos x} \] etc...
 2 years ago

moses_ Group TitleBest ResponseYou've already chosen the best response.0
for multiplying by one like i said above, dont use sin...use something that will reduce everything...
 2 years ago

pottersheep Group TitleBest ResponseYou've already chosen the best response.1
i tried turning tan into sin/cos and cot into cis/sin and then I got all tans and cos's too
 2 years ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

Mertsj Group TitleBest ResponseYou've already chosen the best response.0
dw:1322797409581:dw
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
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 \[\frac{1}{\sin(x)+\cos(x)}=\sin(x)+\cos(x)\] ...
 2 years ago

Mertsj Group TitleBest ResponseYou've already chosen the best response.0
Now. Use the same technique on the right side.
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
this doesn't look like an identity
 2 years ago

moses_ Group TitleBest ResponseYou've already chosen the best response.0
oh my apologies
 2 years ago

pottersheep Group TitleBest ResponseYou've already chosen the best response.1
Wow thanks for all that help...its supposed to be an identity though? sigh....perhaps its teachers mistake
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
it is not
 2 years ago

pottersheep Group TitleBest ResponseYou've already chosen the best response.1
oh wait a minute if I change both sides then I can get them to be equal
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
try x=pi/3 both sides have different outputs
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
one side is approximately 1.366 the other side is .73205
 2 years ago

Mertsj Group TitleBest ResponseYou've already chosen the best response.0
No. It is not. You have shown that they indeed are reciprocals which was apparent by a casual inspection of the original problem.
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
the equation does hold for some values but not all which makes it not an identity
 2 years ago

pottersheep Group TitleBest ResponseYou've already chosen the best response.1
Thanks you guys =) i guess i need much more practice
 2 years ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
when I simplified both sides after making them in terms of sin and cos I got (sinxcosx)(sinx^2cosx^2)  cosx^2*sinx^2 for both the left hand side and the right hand side
 2 years ago

bluebrandon Group TitleBest ResponseYou've already chosen the best response.0
start withdw:1322798230411:dw
 2 years ago

Mertsj Group TitleBest ResponseYou've already chosen the best response.0
So how did you get the denominator to be a product instead of a difference?
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
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
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
so you wouldn't have that product anymore
 2 years ago

pottersheep Group TitleBest ResponseYou've already chosen the best response.1
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....
 2 years ago

pottersheep Group TitleBest ResponseYou've already chosen the best response.1
i still have like 18 questions to go...sigh...
 2 years ago

pottersheep Group TitleBest ResponseYou've already chosen the best response.1
trig identities :P not functions
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
anyways potter we have already shown that this is not an identity do you have any questions as to why we concluded that?
 2 years ago

Mertsj Group TitleBest ResponseYou've already chosen the best response.0
Also, potter, are you absolutely sure that you copied the problem correctly?
 2 years ago

pottersheep Group TitleBest ResponseYou've already chosen the best response.1
Nope I understand...thank you everyone :) now i need to practice it
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
ok have fun
 2 years ago

pottersheep Group TitleBest ResponseYou've already chosen the best response.1
lol you guys are calling me potter hehe. and yeah but our teacher does have a bad reputation of making typos lol
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
is potter okay?
 2 years ago

pottersheep Group TitleBest ResponseYou've already chosen the best response.1
yep potter's fine :)
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
harry potter? lol is that what you were thinking?
 2 years ago

pottersheep Group TitleBest ResponseYou've already chosen the best response.1
yeah xD
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
it wouldn't be so bad to be the most powerful wizard of all time
 2 years ago

pottersheep Group TitleBest ResponseYou've already chosen the best response.1
yeah :( then i woulnt have to worry about these math problems lol
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
you could just turn a math problem you didn't feel like working on into a bunny
 2 years ago

pottersheep Group TitleBest ResponseYou've already chosen the best response.1
LOL hahaha
 2 years ago

pottersheep Group TitleBest ResponseYou've already chosen the best response.1
too bad i cant give u another medal for that :P
 2 years ago
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