VeroZarate
Verify that cot x sec4x = cot x + 2 tan x + tan3x
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VeroZarate
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so i know cotx is cos/sin
VeroZarate
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idk if we can change the tan though to sin/cos?
binarymimic
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this one is hard but try this
binarymimic
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first divide everything by cot x
binarymimic
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\[\sec^{4} x = 1 + 2\frac{ \tan x }{ \cot x } + \frac{ \tan^{3} x }{ \cot x }\]
binarymimic
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and we know that
\[\frac{ \tan x }{ \cot x } = \frac{ \frac{ \sin x }{ \cos x } }{ \frac{ \cos x }{ \sin x } } = \frac{\sin^{2} x }{ \cos^{2} x } = \tan^{2} x\]
VeroZarate
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would it apply to the second one but it would be (sin^3x/cos^3x)/cosx/sinx ?
binarymimic
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that would simplify it to:
\[\sec^{4} x = 1 + 2 \tan^{2} x + \tan^{4} x\]
binarymimic
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yeah the second one is:
\[\frac{ \tan^{3} x }{ \cot x } = \frac{ \tan x }{ \cot x } * \tan^{2} x = \tan^{2} x * \tan^{2} x = \tan^{4} x\]
VeroZarate
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where did you get the second tan^2x
binarymimic
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I divided tan^3 x into tan^x * tan^2 x
binarymimic
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|dw:1357511455432:dw|
VeroZarate
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oh ok
binarymimic
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So this is why we have the following equation so far:
\[\sec^{4} x = 1 + 2 \tan^{2} x + \tan^{4} x\]
binarymimic
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next step is to subtract both sides by tan^4 x
binarymimic
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\[\sec^{4} x - \tan^{4} x = 1 + 2 \tan^{2} x\]
VeroZarate
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and divide both side by2?
binarymimic
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no there is a trick here now. notice the left side is the difference of two squares
binarymimic
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\[\sec^{4} x - \tan^{4} x = (\sec^{2} x)^{2} - (\tan^{2} x)^{2}\]
binarymimic
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remember that a^2 - b^2 = (a - b)(a + b) so we can now write the left side as:
\[\sec^{4} x - \tan^{4} x = (\sec^{2} x - \tan^{2} x)(\sec^{2} x + \tan^{2} x)\]
VeroZarate
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right you foil it?
binarymimic
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we kind of did the reverse of FOIL, we factored it
binarymimic
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So now we have:
\[(\sec^{2} x - \tan^{2} x)(\sec^{2} x + \tan^{2} x) = 1 + 2 \tan^{2} x\]
binarymimic
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but do you remember the identity
sec^2 x = tan^2 x + 1
VeroZarate
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yes but there is the 2?
binarymimic
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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
binarymimic
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Then we see how
binarymimic
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\[(\sec^{2} x - \tan^{2} x)(\sec^{2} x + \tan^{2} x) = 1 + 2 \tan^{2} x\]
becomes
\[(1)(\sec^{2} x + \tan^{2} x) = 1 + 2 \tan^2{x}\]
VeroZarate
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isnt it sec^4
binarymimic
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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
VeroZarate
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ok so when you subtract tan^2x?
binarymimic
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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:
\[(\sec^{2} x - \tan^{2} x)(\sec^{2} x + \tan^{2} x) = (1)(\sec^{2} x + \tan^{2} x) \]
binarymimic
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so we are now left with
\[\sec^2{x} + \tan^{2} x = 1 + 2 \tan^{2} x\]
VeroZarate
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ok you had lost me. now i get it sorry
binarymimic
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its a complicated problem compared to the last one o_o
VeroZarate
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yes it is. i still have to do 5 more assignments. lol
binarymimic
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so now we have
\[\sec^2{x} + \tan^{2} x = 1 + 2 \tan^{2} x\]
can you see the next step ?
VeroZarate
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substitute the sec^2x with 1+ tan^2x?
binarymimic
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we could, or we could save a step and just subtract tan^2 x from both sides
binarymimic
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if we subtract tan^2 x from both sides we get
\[\sec^{2} x = 1 + \tan^{2} x\]
binarymimic
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which is an identity
VeroZarate
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ok
VeroZarate
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but how does this equal to cotxsec^4x?
binarymimic
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the whole idea is to reduce the original equation into a trig identity
binarymimic
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if you follow the steps then that's precisely what we did
VeroZarate
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to verify that it equals on both sides?
binarymimic
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if you manipulate an equation using algebra then whatever you end up with is equal to what you started with.
VeroZarate
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oh ok thanks so much!
binarymimic
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np