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VeroZarateBest ResponseYou've already chosen the best response.0
so i know cotx is cos/sin
 one year ago

VeroZarateBest ResponseYou've already chosen the best response.0
idk if we can change the tan though to sin/cos?
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
this one is hard but try this
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
first divide everything by cot x
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
\[\sec^{4} x = 1 + 2\frac{ \tan x }{ \cot x } + \frac{ \tan^{3} x }{ \cot x }\]
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
and we know that \[\frac{ \tan x }{ \cot x } = \frac{ \frac{ \sin x }{ \cos x } }{ \frac{ \cos x }{ \sin x } } = \frac{\sin^{2} x }{ \cos^{2} x } = \tan^{2} x\]
 one year ago

VeroZarateBest ResponseYou've already chosen the best response.0
would it apply to the second one but it would be (sin^3x/cos^3x)/cosx/sinx ?
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
that would simplify it to: \[\sec^{4} x = 1 + 2 \tan^{2} x + \tan^{4} x\]
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
yeah the second one is: \[\frac{ \tan^{3} x }{ \cot x } = \frac{ \tan x }{ \cot x } * \tan^{2} x = \tan^{2} x * \tan^{2} x = \tan^{4} x\]
 one year ago

VeroZarateBest ResponseYou've already chosen the best response.0
where did you get the second tan^2x
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
I divided tan^3 x into tan^x * tan^2 x
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
dw:1357511455432:dw
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
So this is why we have the following equation so far: \[\sec^{4} x = 1 + 2 \tan^{2} x + \tan^{4} x\]
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
next step is to subtract both sides by tan^4 x
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
\[\sec^{4} x  \tan^{4} x = 1 + 2 \tan^{2} x\]
 one year ago

VeroZarateBest ResponseYou've already chosen the best response.0
and divide both side by2?
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
no there is a trick here now. notice the left side is the difference of two squares
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
\[\sec^{4} x  \tan^{4} x = (\sec^{2} x)^{2}  (\tan^{2} x)^{2}\]
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
remember that a^2  b^2 = (a  b)(a + b) so we can now write the left side as: \[\sec^{4} x  \tan^{4} x = (\sec^{2} x  \tan^{2} x)(\sec^{2} x + \tan^{2} x)\]
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
we kind of did the reverse of FOIL, we factored it
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
So now we have: \[(\sec^{2} x  \tan^{2} x)(\sec^{2} x + \tan^{2} x) = 1 + 2 \tan^{2} x\]
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
but do you remember the identity sec^2 x = tan^2 x + 1
 one year ago

VeroZarateBest ResponseYou've already chosen the best response.0
yes but there is the 2?
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
we'll get to that :) so if sec^2 x = tan^2 x + 1 the if we subtract tan^2 from both sides we get sec^2 x  tan^2 x = 1
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
\[(\sec^{2} x  \tan^{2} x)(\sec^{2} x + \tan^{2} x) = 1 + 2 \tan^{2} x\] becomes \[(1)(\sec^{2} x + \tan^{2} x) = 1 + 2 \tan^2{x}\]
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
No we factored \[\sec^{4} x  \tan^{4} x \] into \[(\sec^{2} x  \tan^{2} x)(\sec^{2} x + \tan^{2} x)\] because it is a difference of sqaures
 one year ago

VeroZarateBest ResponseYou've already chosen the best response.0
ok so when you subtract tan^2x?
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
There is a trig identity: \[\sec^{2} x = 1 + \tan^{2} x\] i am saying, if we rewrite it as: \[\sec^{2} x  \tan^{2} x = 1\] then the left factor on our left hand side reduces to just 1: \[(\sec^{2} x  \tan^{2} x)(\sec^{2} x + \tan^{2} x) = (1)(\sec^{2} x + \tan^{2} x) \]
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
so we are now left with \[\sec^2{x} + \tan^{2} x = 1 + 2 \tan^{2} x\]
 one year ago

VeroZarateBest ResponseYou've already chosen the best response.0
ok you had lost me. now i get it sorry
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
its a complicated problem compared to the last one o_o
 one year ago

VeroZarateBest ResponseYou've already chosen the best response.0
yes it is. i still have to do 5 more assignments. lol
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
so now we have \[\sec^2{x} + \tan^{2} x = 1 + 2 \tan^{2} x\] can you see the next step ?
 one year ago

VeroZarateBest ResponseYou've already chosen the best response.0
substitute the sec^2x with 1+ tan^2x?
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
we could, or we could save a step and just subtract tan^2 x from both sides
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
if we subtract tan^2 x from both sides we get \[\sec^{2} x = 1 + \tan^{2} x\]
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
which is an identity
 one year ago

VeroZarateBest ResponseYou've already chosen the best response.0
but how does this equal to cotxsec^4x?
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
the whole idea is to reduce the original equation into a trig identity
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
if you follow the steps then that's precisely what we did
 one year ago

VeroZarateBest ResponseYou've already chosen the best response.0
to verify that it equals on both sides?
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
if you manipulate an equation using algebra then whatever you end up with is equal to what you started with.
 one year ago

VeroZarateBest ResponseYou've already chosen the best response.0
oh ok thanks so much!
 one year ago
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