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anonymous
 3 years ago
Prove:
anonymous
 3 years ago
Prove:

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[Tan^2x  \cos^2x = 1/\cos^2  1  \cos^2x\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you rewrite that right side? is that \[\frac 1{cos^2 x  1}  \cos ^2 x\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352207602501:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you know this identity right? \[\sin^2 x = 1  \cos^2 x\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait..nevermind what i said....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac 1{\cos^2x}  1  \cos^2 x\] convert to common denominators first \[\implies \frac 1{\cos^2 x}  \frac{\cos^2 x}{\cos^2 x}  \frac{\cos^4 x}{\cos^2 x}\] agree?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes :) I was on the right track whew :P

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now combine the numerators \[\frac{1  \cos^2 x  \cos ^ 4 x}{\cos^2 x}\] is this what you did?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes but Instead if making it cos^4x I cancelled out the top bracket Cos^2x with the bottom one...which is probably where I went wrong?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no. that works too \[\frac{1  \cos^2 x}{\cos^2 x}  \cos^2 x\] yes?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm, why didnt you just totally cancel out the denominator?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I was left with 1cos^2xcos^2x :S

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352208030577:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm whats the reason I couldnt?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0observe what i did earlier \[\frac 1{\cos^2 x}  1  \cos^2 x \implies \frac 1{\cos^2 x}  \frac{\cos^2 x}{\cos^2 x}  \frac{\cos^4 x}{\cos^2 x}\] each term has its own denominator

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1352208100756:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ap calc etcher tell me that

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if you cancel out the cos^2 x below cos^4 x, you can only do it there, and not in the other terms. make sense?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so anyway, the RHS simplifies to \[\frac{1  \cos^2 x}{\cos^2 x}  \cos^2 x\] i assume you know how to proceed form tehre?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0should i move to the left side?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wo think i got it? L.S) will equel 1cos^2/cos^2x  Cos^2x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ls = Rs Sweet! Thanks for your help! I also found a few more I am having trouble with...this next question has a tan^3x which I haven't used in a while i forget what i can do with it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0tan^3 x = tan^2 x * tan x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or maybe tan^3 x = sin^3 x / cos^3 x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\tan ^{2}x  \cos^{2}x =\frac{1}{ \cos^2(x)} 1 \cos^{2}x \] \[\tan ^{2}x  \cos^{2}x =\frac{(1\cos^{2}x) }{ \cos^2(x)}  \cos^{2}x \] which is \\[\tan ^{2}x =\frac{(1\cos^{2}x) }{ \cos^2(x)} \] which is \[\tan ^{2}x =\frac{(\sin^{2}x) }{ \cos^2(x)}\] which satisfis the relation

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks guys, lemme try this other one and i cant get it ill post a new question
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