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

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

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

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

pottersheep
 3 years ago
Best ResponseYou've already chosen the best response.1Oh....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 :(

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

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

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

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

bluebrandon
 3 years ago
Best ResponseYou've already chosen the best response.0try 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

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.3i'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
 3 years ago
Best ResponseYou've already chosen the best response.0Now. Use the same technique on the right side.

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

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

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

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

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

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

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

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

bluebrandon
 3 years ago
Best ResponseYou've already chosen the best response.0when 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

bluebrandon
 3 years ago
Best ResponseYou've already chosen the best response.0start withdw:1322798230411:dw

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

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

pottersheep
 3 years ago
Best ResponseYou've already chosen the best response.1omg! 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
 3 years ago
Best ResponseYou've already chosen the best response.1i still have like 18 questions to go...sigh...

pottersheep
 3 years ago
Best ResponseYou've already chosen the best response.1trig identities :P not functions

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

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

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

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

pottersheep
 3 years ago
Best ResponseYou've already chosen the best response.1yep potter's fine :)

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

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

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

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

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