cot x sec4x = cot x + 2 tan x + tan3x
I have to make the left side look like the right to prove this equation is true. I've done most of it but I'm stuck at the end
cot x sec^4x = cot x + 2 tan x + tan^3x
cot x sec^2x sec^2x = cot x + 2 tan x + tan^3x
cot x + 1 + tan^2 x + 1 + tan^2 x = cot x + 2 tan x + tan^3x

- Helpjebal

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- Helpjebal

cot x + 2 + tan^2 x +tan^2 x

- Helpjebal

sorry the right side has to look like the left side

- Helpjebal

now i'm really confused....

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## More answers

- anonymous

this might help you: cos2x = 2cos^2(x) -1

- anonymous

it doesnt matter which side should look like which one

- anonymous

as long as you can prove that you can derive one side from the other, your proof is correct

- mathmale

Yes, and you might also think about relevant trig identities:
cot x = (cos x) / (sin x)
sec x = 1 / (cos x)
These are just examples. You might also try tan x = (sin x) / (cos x)

- anonymous

i'd say start with the right side. it'll be easier

- anonymous

left*

- Helpjebal

can I just do this?
cot x + 2 + tan x + tan^3 x
cot x + 2 + tan^4 x
cot x + 2(1 + tan^2 x)
cot x + 2(sec^2 x)
cot x +sec ^4 x
which would just be cot x sec ^4 x like in the original?

- anonymous

is it sec^4x or sec4x in the left side?

- mathmale

@Helpjebal: your proposed expression shows ' 2 ' standing by itself. This is not the case in the given, original, equation. So, unfortunately, my answer is "no."

- Helpjebal

it's sec^4x on the left side

- anonymous

and how did you get tan^3x from tan3x?

- anonymous

oh

- mathmale

You might have to experiment (e. g., start with the left side, start with the right side) until you find yourself on the right track, especially if you're not used to this sort of problem.
Again I urge you to consider the trig identities I typed to you. So far I haven't seen you using them.

- Helpjebal

I didn't write it properly, it's cot x sec^4x = cot x + 2 tan x + tan^3x

- mathmale

I agree that you can work on either side, or on both sides. If you change the right side, you can always change it back to its original form in the end.
The key to proving this identity lies in the proper use fo trig identities. What is the common trig identity that shows the relationship between the cot and the sec functions?

- mathmale

Hint: this identity stems from the even more common\[\sin^2x+\cos^2x=1\]

- Helpjebal

okay so I think I know how to use this trig identities you showed earlier...
cot x + 2 tan x + tan^3 x
1/ tan x + 2 tan x + tan^3 x
(1 / tan x) + (2 tan x + tan^3 x / 1)
1 + 2 tan x + tan^3 x / 1 tan x
that's where I've gotten... but you said that sin^2x+cos^2x=1
so I could also do
cot x + 2 tan x + tan^3 x
cos(x)/sin(x) + 2(sin(x)/cos(x)) + tan^3x?

- mathmale

The trick here is to reduce the number of different trig functions in your expressions. Of course you can define tan x as (sin x) / (cos x), but do you really want to introduce sin x and cos x here, when the primary trig functions of the given expression are cot x and sec x? I think not.

- mathmale

Have y ou a table of trig identities available? If so, review \[\sin^2x + \cos^2x = 1\]

- mathmale

and find the identity involving \[\cot^2x\]

- mathmale

Then apply this "new" identity to simplifying the given expression. Again: seek to REDUCE (not increase) the number of trig functions you're using.

- anonymous

ok i am back

- anonymous

i did the question. there are only three steps to solving it

- anonymous

first, you want to convert the sec^4x to something with a lower order

- anonymous

use the identity sec^x = 1+tan^2x

- anonymous

sec^2x = 1+tan^2x

- Helpjebal

Thats how I was originally trying to solve it
cot x + 1 + tan^2x + 1 + tan^2x
cot x + 2 + tan^4 x
but I got incredibly stuck...

- Helpjebal

The only identity i know of involving cot^2 x is + cot^2x = csc^2x

- mathmale

Please double check before you make substitutions such as "cos x + 1." That is NOT an identity.

- Helpjebal

I never said cos x + 1...
2(1 + tan x) is the same as 1 + tan^2x + 1 + tan^2x thats why i wrote it like that

- mathmale

Check out physo's suggestion: sec^x = 1+tan^2x. Actually, this should be
(sec x)^2 = 1 + (tan x)^2. Is it correct? Double check. If yes, apply it to simplifying the given expression.

- anonymous

i'll give you the first step to help you:\[cotx(\sec^2x)^2\]

- mathmale

@Helpjebal :
You did type in the following:
cot x + 1 + tan^2x + 1 + tan^2x
cot x + 2 + tan^4 x

- anonymous

now use the identity on sec^x and expand it

- anonymous

sec^2x

- mathmale

Let's let physo help here. I'll be in the background if you need me.

- Helpjebal

okay so physo we're on the left side right?
so...
cot x sec^4x
cot x (sec^2 x)^2
cot x (1 + tan^2x)^2
am I on the right path?

- anonymous

yes

- anonymous

do you remember (a+b)^2 = a^2 + 2ab + b^2?

- Helpjebal

yes I remember that so...
cot x sec^4x
cot x (sec^2 x)^2
cot x (1 + tan^2x)^2
cot x + 1 + 2 tan^2 x + tan^4x

- anonymous

you need to take the brackets into consideration

- mathmale

this is fine:
cot x sec^4x
cot x (sec^2 x)^2
cot x (1 + tan^2x)^2
but the next line is not.

- anonymous

yup

- mathmale

physo is right: those parentheses / brackets are very important gatekeepers. We can't just ignore them. Anything within ( ) must be done first, followed by any exponentiation.

- Helpjebal

okay so
(1 + tan^2x)(1 + tan^2x)
1 + tan^2x + tan^2x + 2 tan^2x
1 + tan^4x + 2 tan^2x
making that correction it would be
cot x sec^4x
cot x (sec^2 x)^2
cot x (1 + tan^2x)^2
cot x + 1 + tan^4x + 2 tan^2x

- Helpjebal

wait sorry

- Helpjebal

i did that wrong again didn't i...

- mathmale

Please note: because that cot x is OUTSIDE your parentheses, you must multiply everything INSIDE the parentheses by cot x. Distributive rule of multiplication. ;(

- anonymous

yes mathmale is correct. you squared it correctly but you forgot to multiply it with cotx

- Helpjebal

ohhhhhh
cot x (1 + tan^4x + 2 tan^2x)
cot x + tan^3 + 2 tan x

- anonymous

there you go

- mathmale

Hold it, physo. That wasn't my work.

- Helpjebal

Thank you guys so much! I'm so incredibly thankful!

- mathmale

So glad you've been able to solve this problem!

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