Show(1-tanx)/(1+tanx)^2 = cos2x?

- anonymous

Show(1-tanx)/(1+tanx)^2 = cos2x?

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

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

- anonymous

cos(2x)

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

To me, it is not an identity. Counterexample
Let test x = pi/6,
the LHS = 0.1698729811
while the RHS =0.5

- anonymous

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

- anonymous

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

- xapproachesinfinity

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

- xapproachesinfinity

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

- anonymous

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

- anonymous

?

- xapproachesinfinity

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

- xapproachesinfinity

manupilate that a bit more into sin and cos

- xapproachesinfinity

then after that do the same with top too

- anonymous

but there is the square2 there confusing me

- xapproachesinfinity

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

- anonymous

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

- xapproachesinfinity

no

- xapproachesinfinity

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

- xapproachesinfinity

let me start a bit
\(\huge \frac{1-\tan x}{\sec^2x +2\frac{\sin x}{\cos x}}=\frac{\frac{\cos x-\sin x}{\cos x}}{\frac{1}{\cos^2x}+2\frac{\sin x}{\cos x }}\)

- xapproachesinfinity

see if you can finish it off

- xapproachesinfinity

you can cancel that cos that is in top and bottom

- IrishBoy123

|dw:1440019023761:dw|

- anonymous

|dw:1440019515190:dw|

- xapproachesinfinity

damn why did post the wrong question lol

- xapproachesinfinity

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

- anonymous

haha sorry but i still learn to u

- xapproachesinfinity

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

- xapproachesinfinity

work the other way around from cos2x

- IrishBoy123

|dw:1440019749527:dw|

- xapproachesinfinity

cos2x=cos^2x-sin^2x

- xapproachesinfinity

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

- xapproachesinfinity

that's one step

- xapproachesinfinity

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

- xapproachesinfinity

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

- xapproachesinfinity

should be only 1+tan^2x

- anonymous

hh ya realy sure

- anonymous

i need to proven the answer is cos2x

- xapproachesinfinity

|dw:1440020130101:dw|

- anonymous

yup

- xapproachesinfinity

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

- xapproachesinfinity

it cannot be right the way you wrote it please make sure
post the picture yo getting u all confused now hhhh

- anonymous

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

- xapproachesinfinity

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

- xapproachesinfinity

unless you have a different question

- anonymous

the last picture you post is my question

- IrishBoy123

screenshot, or url

- IrishBoy123

if i had a penny....

- xapproachesinfinity

please screenshot that thing man we got confused now

- xapproachesinfinity

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

- anonymous

ya this is my question as exam

- freckles

does it say to disprove or to prove?

- freckles

or does it just say to show?

- xapproachesinfinity

well depends on what the question really states
perhaps it want to you see whether that is an identity or not

- xapproachesinfinity

so to answer this it is not true
not an identity

- anonymous

|dw:1440020646681:dw|

- anonymous

if like this?

- xapproachesinfinity

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

- anonymous

so sad

- IrishBoy123

|dw:1440020700278:dw|

- anonymous

lrisboy 123 i think there is no number inn answer
because cos2x is the answer
jz need to prove it (1-tan^2x)/(1+tan^2x)^2 equal to cos2x

- IrishBoy123

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

- freckles

but they already said it cannot be proven
because it is not an identity
@IrishBoy123 was giving a counterexample aka proving it is a false equation

- IrishBoy123

thanks @freckles

- freckles

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

- freckles

because this is a true equation

- freckles

on the left hand side's domain of course

- anonymous

\(\dfrac{1-tan^2(x)}{(1+tan^2(x))^2}\\=\dfrac{1-tan^2(x)}{sec^4(x) }\\= \dfrac{1}{sec^4x}-\dfrac{tan^2x}{sec^4x}\\= cos^4x-cos^2xsin^2x\\=cos^2(x)(cos^2(x)-sin^2(x) )= cos^2(x) (cos(2x))\)
No way for you to get cos (2x) only.

Looking for something else?

Not the answer you are looking for? Search for more explanations.