anonymous
  • anonymous
Show(1-tanx)/(1+tanx)^2 = cos2x?
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
is that cos^2(x) or cos(2x)
anonymous
  • anonymous
cos(2x)
anonymous
  • anonymous
okay start with the trig identiy for cos(2x)

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anonymous
  • anonymous
http://prntscr.com/86l5g2 this one
anonymous
  • anonymous
turn tan in sin/cos and play around with it, tell me if u are still stuck
anonymous
  • anonymous
how to settle the(1+tanx)^2 i keep cant not get the answer cos2x
anonymous
  • anonymous
To me, it is not an identity. Counterexample Let test x = pi/6, the LHS = 0.1698729811 while the RHS =0.5
anonymous
  • anonymous
can do full step to me? i realy no idea==
anonymous
  • anonymous
to convert (1-tanx)/(1+tanx)^2 to cos2x
xapproachesinfinity
  • xapproachesinfinity
try to simplify (1+tanx)^2 to (sec^2x+2tanx) sec^2x+2sinx/cosx
xapproachesinfinity
  • xapproachesinfinity
do like dan suggested transfer tan to sin/cos that will be helpful
anonymous
  • anonymous
if it is (1-tan^2x)/(1+tan^2x)^2
anonymous
  • anonymous
?
xapproachesinfinity
  • xapproachesinfinity
hmm no i was talking about the denominator of left hand side (1+tanx)^2
xapproachesinfinity
  • xapproachesinfinity
manupilate that a bit more into sin and cos
xapproachesinfinity
  • xapproachesinfinity
then after that do the same with top too
anonymous
  • anonymous
but there is the square2 there confusing me
xapproachesinfinity
  • xapproachesinfinity
the square you need to distribute like you do with (a+b)^2
anonymous
  • anonymous
is(1+tan^4x) equal to sec^4x?
xapproachesinfinity
  • xapproachesinfinity
no
xapproachesinfinity
  • xapproachesinfinity
there is no such identity we have 1+tan^2x=sec^2x
xapproachesinfinity
  • 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
  • xapproachesinfinity
see if you can finish it off
xapproachesinfinity
  • xapproachesinfinity
you can cancel that cos that is in top and bottom
IrishBoy123
  • IrishBoy123
|dw:1440019023761:dw|
anonymous
  • anonymous
|dw:1440019515190:dw|
xapproachesinfinity
  • xapproachesinfinity
damn why did post the wrong question lol
xapproachesinfinity
  • xapproachesinfinity
i was wondering why your talking about 1-tan^2x in a reply so that was the question
anonymous
  • anonymous
haha sorry but i still learn to u
xapproachesinfinity
  • xapproachesinfinity
well you had us work the wrong thing all the way hhh but no worries
xapproachesinfinity
  • xapproachesinfinity
work the other way around from cos2x
IrishBoy123
  • IrishBoy123
|dw:1440019749527:dw|
xapproachesinfinity
  • xapproachesinfinity
cos2x=cos^2x-sin^2x
xapproachesinfinity
  • xapproachesinfinity
factor cos^2x to get cos^2x(1-tan^2x)
xapproachesinfinity
  • xapproachesinfinity
that's one step
xapproachesinfinity
  • xapproachesinfinity
recall that cos^2x=1/sec^2x
xapproachesinfinity
  • xapproachesinfinity
but again you sure it is (1+tan^2x)^2? make sure
xapproachesinfinity
  • xapproachesinfinity
should be only 1+tan^2x
anonymous
  • anonymous
hh ya realy sure
anonymous
  • anonymous
i need to proven the answer is cos2x
xapproachesinfinity
  • xapproachesinfinity
|dw:1440020130101:dw|
anonymous
  • anonymous
yup
xapproachesinfinity
  • xapproachesinfinity
i got 1-tan^2x/1+tan^2x=cos^2 not square the bottom
xapproachesinfinity
  • xapproachesinfinity
it cannot be right the way you wrote it please make sure post the picture yo getting u all confused now hhhh
anonymous
  • anonymous
but still can get the answer as cos2x jz dono the way how to do
xapproachesinfinity
  • xapproachesinfinity
no it can't if my result is correct that cannot be true
xapproachesinfinity
  • xapproachesinfinity
unless you have a different question
anonymous
  • anonymous
the last picture you post is my question
IrishBoy123
  • IrishBoy123
screenshot, or url
IrishBoy123
  • IrishBoy123
if i had a penny....
xapproachesinfinity
  • xapproachesinfinity
please screenshot that thing man we got confused now
xapproachesinfinity
  • xapproachesinfinity
|dw:1440020383353:dw| if you are talking about this the question is wrong not an identity
anonymous
  • anonymous
ya this is my question as exam
freckles
  • freckles
does it say to disprove or to prove?
freckles
  • freckles
or does it just say to show?
xapproachesinfinity
  • xapproachesinfinity
well depends on what the question really states perhaps it want to you see whether that is an identity or not
xapproachesinfinity
  • xapproachesinfinity
so to answer this it is not true not an identity
anonymous
  • anonymous
|dw:1440020646681:dw|
anonymous
  • anonymous
if like this?
xapproachesinfinity
  • xapproachesinfinity
look if it is not an identity no matter what you try you won't get there haha
anonymous
  • anonymous
so sad
IrishBoy123
  • IrishBoy123
|dw:1440020700278:dw|
anonymous
  • 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
  • IrishBoy123
there should be *every * number in answer, if it was true. yes? x = 0 works. !!
freckles
  • 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
  • IrishBoy123
thanks @freckles
freckles
  • freckles
if you meant \[\frac{1-\tan^2(x)}{1+\tan^2(x)}=\cos(2x) \] we would have had more luck
freckles
  • freckles
because this is a true equation
freckles
  • freckles
on the left hand side's domain of course
anonymous
  • 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.

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