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is that cos^2(x) or cos(2x)

cos(2x)

okay start with the trig identiy for cos(2x)

http://prntscr.com/86l5g2 this one

turn tan in sin/cos and play around with it, tell me if u are still stuck

how to settle the(1+tanx)^2
i keep cant not get the answer cos2x

can do full step to me?
i realy no idea==

to convert (1-tanx)/(1+tanx)^2 to cos2x

try to simplify (1+tanx)^2 to (sec^2x+2tanx)
sec^2x+2sinx/cosx

do like dan suggested transfer tan to sin/cos
that will be helpful

if it is (1-tan^2x)/(1+tan^2x)^2

hmm no i was talking about the denominator of left hand side
(1+tanx)^2

manupilate that a bit more into sin and cos

then after that do the same with top too

but there is the square2 there confusing me

the square you need to distribute
like you do with (a+b)^2

is(1+tan^4x) equal to sec^4x?

there is no such identity
we have 1+tan^2x=sec^2x

see if you can finish it off

you can cancel that cos that is in top and bottom

|dw:1440019023761:dw|

|dw:1440019515190:dw|

damn why did post the wrong question lol

i was wondering why your talking about 1-tan^2x in a reply so that was the question

haha sorry but i still learn to u

well you had us work the wrong thing all the way hhh
but no worries

work the other way around from cos2x

|dw:1440019749527:dw|

cos2x=cos^2x-sin^2x

factor cos^2x
to get cos^2x(1-tan^2x)

that's one step

recall that cos^2x=1/sec^2x

but again you sure it is (1+tan^2x)^2? make sure

should be only 1+tan^2x

hh ya realy sure

i need to proven the answer is cos2x

|dw:1440020130101:dw|

yup

i got 1-tan^2x/1+tan^2x=cos^2
not square the bottom

but still can get the answer as cos2x
jz dono the way how to do

no it can't if my result is correct that cannot be true

unless you have a different question

the last picture you post is my question

screenshot, or url

if i had a penny....

please screenshot that thing man we got confused now

|dw:1440020383353:dw|
if you are talking about this the question is wrong not an identity

ya this is my question as exam

does it say to disprove or to prove?

or does it just say to show?

so to answer this it is not true
not an identity

|dw:1440020646681:dw|

if like this?

look if it is not an identity no matter what you try you won't get there haha

so sad

|dw:1440020700278:dw|

there should be *every * number in answer, if it was true. yes?
x = 0 works. !!

thanks @freckles

if you meant
\[\frac{1-\tan^2(x)}{1+\tan^2(x)}=\cos(2x) \]
we would have had more luck

because this is a true equation

on the left hand side's domain of course