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cot x + 2 + tan^2 x +tan^2 x
sorry the right side has to look like the left side
now i'm really confused....
this might help you: cos2x = 2cos^2(x) -1
it doesnt matter which side should look like which one
as long as you can prove that you can derive one side from the other, your proof is correct
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)
i'd say start with the right side. it'll be easier
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?
is it sec^4x or sec4x in the left side?
@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."
it's sec^4x on the left side
and how did you get tan^3x from tan3x?
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.
I didn't write it properly, it's cot x sec^4x = cot x + 2 tan x + tan^3x
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?
Hint: this identity stems from the even more common\[\sin^2x+\cos^2x=1\]
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?
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.
Have y ou a table of trig identities available? If so, review \[\sin^2x + \cos^2x = 1\]
and find the identity involving \[\cot^2x\]
Then apply this "new" identity to simplifying the given expression. Again: seek to REDUCE (not increase) the number of trig functions you're using.
ok i am back
i did the question. there are only three steps to solving it
first, you want to convert the sec^4x to something with a lower order
use the identity sec^x = 1+tan^2x
sec^2x = 1+tan^2x
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...
The only identity i know of involving cot^2 x is + cot^2x = csc^2x
Please double check before you make substitutions such as "cos x + 1." That is NOT an identity.
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
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.
i'll give you the first step to help you:\[cotx(\sec^2x)^2\]
@Helpjebal : You did type in the following: cot x + 1 + tan^2x + 1 + tan^2x cot x + 2 + tan^4 x
now use the identity on sec^x and expand it
Let's let physo help here. I'll be in the background if you need me.
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?
do you remember (a+b)^2 = a^2 + 2ab + b^2?
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
you need to take the brackets into consideration
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.
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.
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
i did that wrong again didn't i...
Please note: because that cot x is OUTSIDE your parentheses, you must multiply everything INSIDE the parentheses by cot x. Distributive rule of multiplication. ;(
yes mathmale is correct. you squared it correctly but you forgot to multiply it with cotx
ohhhhhh cot x (1 + tan^4x + 2 tan^2x) cot x + tan^3 + 2 tan x
there you go
Hold it, physo. That wasn't my work.
Thank you guys so much! I'm so incredibly thankful!
So glad you've been able to solve this problem!