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Verify that cot x sec4x = cot x + 2 tan x + tan3x

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so i know cotx is cos/sin
idk if we can change the tan though to sin/cos?
this one is hard but try this

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Other answers:

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

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