anonymous
  • anonymous
Verify that cot x sec4x = cot x + 2 tan x + tan3x
Trigonometry
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
so i know cotx is cos/sin
anonymous
  • anonymous
idk if we can change the tan though to sin/cos?
anonymous
  • anonymous
this one is hard but try this

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

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