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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

first divide everything by cot x

\[\sec^{4} x = 1 + 2\frac{ \tan x }{ \cot x } + \frac{ \tan^{3} x }{ \cot x }\]

would it apply to the second one but it would be (sin^3x/cos^3x)/cosx/sinx ?

that would simplify it to:
\[\sec^{4} x = 1 + 2 \tan^{2} x + \tan^{4} x\]

where did you get the second tan^2x

I divided tan^3 x into tan^x * tan^2 x

|dw:1357511455432:dw|

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

\[\sec^{4} x - \tan^{4} x = (\sec^{2} x)^{2} - (\tan^{2} x)^{2}\]

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?

Then we see how

isnt it sec^4

ok so when you subtract tan^2x?

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

ok

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?

oh ok thanks so much!

np