Question

Substitute identities on the left to match the right hand side of the equation.

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

Answer 1

so the equation is

\[\large \cot(x)\sec(4x) = \cot(x) + 2\tan(x) + \tan(3x)\]

right?

Answer 2

ugh not quite let me use the wonderful tools this site provide to make it easier

Answer 3

cot(x) sec^4x = cot(x) + 2tan (x) + tan^3x

Answer 4

except that I couldn't figure out how to use the tools so I did that instead

Answer 5

oh ok, so it's

\[\large \cot(x)\sec^4(x) = \cot(x) + 2\tan(x) + \tan^3(x)\]

Answer 6

one sec while I think on how to do this one

Answer 7

here does this help?
file:///C:/Users/IBMT61/Pictures/05.03%20Proving%20Trigonometric%20Equations.htm

Answer 8

sure take your time

Answer 9

no I can't access that link at all (it's local to your computer)

Answer 10

it's fine though, I can figure it out

Answer 11

hmm let me try this then...

Answer 12

am I a genius in a small form? :D

Answer 13

oh that works much better, thanks

Answer 14

The English nerd doesn't break the tech department for once...praise the gods

Answer 15

Isn't it worded really strangely though? If it wasn't for the fact that I KNOW I have to chance the LEFT side I'd be confused and having to wait for my teacher to ask that

Answer 16

change*

Answer 17

well the instructions are worded a bit oddly

Answer 18

but normally you start with the more complicated side and you simplify it with identities to change it to the other side you didn't pick

Answer 19

This is all that the lesson has...everything else is in a generic online college text book...my high school lessons page itself however only offers THIS

Answer 20

Here's how I did it. I started on the right side (because it was much more complicated) and simplified it to get the left side. I did NOT alter the left side at all.

Answer 21

\[\large \cot(x)\sec^4(x) = \cot(x) + 2\tan(x) + \tan^3(x)\]

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

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

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

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

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

\[\large \cot(x)\sec^4(x) = \frac{(\sec^2(x))^2}{\tan(x)}\]

\[\large \cot(x)\sec^4(x) = \frac{\sec^4(x)}{\tan(x)}\]

\[\large \cot(x)\sec^4(x) = \frac{1}{\tan(x)}*\sec^4(x)\]

\[\large \cot(x)\sec^4(x) = \cot(x)\sec^4(x)\]

Answer 22

The thing is that my course specifically states that I have to change the left side...as you can see in that png file I sent you last...I'm sure that means they want me to do things in a much harder way...of bloody course they do

Answer 23

hmm, there's probably a way to do that, let me think

Answer 24

oh duh, you just follow my work backwards lol

Answer 25

here let me see if I can't translate what I mean

Answer 26

Merci beaucoup

Answer 27

\[\large \cot(x)\sec^4(x) = \cot(x) + 2\tan(x) + \tan^3(x)\]

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

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

\[\large \frac{(\sec^2(x))^2}{\tan(x)} = \cot(x) + 2\tan(x) + \tan^3(x)\]

\[\large \frac{(1+\tan^2(x))^2}{\tan(x)} = \cot(x) + 2\tan(x) + \tan^3(x)\]

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

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

\[\large \frac{1}{\tan(x)} + 2\tan(x) + \tan^3(x) = \cot(x) + 2\tan(x) + \tan^3(x)\]

\[\large \cot(x) + 2\tan(x) + \tan^3(x) = \cot(x) + 2\tan(x) + \tan^3(x)\]



The identity is verified.

Answer 28

oh btw that "duh" was just me thinking out loud and me thinking "oh yeah I should have known that", it wasn't aimed at you (I'm sure you know that though)

anyways, i hope that all makes sense

Answer 29

I gotta read through that and see if I get to where your mind is...but I'll let you know if I feel stuck at all...and thank you for this! I wish I could give you a medal but I don't know how :(

Answer 30

you just click "best response" but I don't mind either way

Answer 31

and yeah, feel free to ask any questions you have about any of that above