SOLUTION: Verify this identity : (tan^2(x)-1)/(1+tan^2(x)) = 1-2cos^2(x) I've started a couple different options but none are working out for me. Here is what I have so far : A) mu

Algebra ->  Trigonometry-basics -> SOLUTION: Verify this identity : (tan^2(x)-1)/(1+tan^2(x)) = 1-2cos^2(x) I've started a couple different options but none are working out for me. Here is what I have so far : A) mu      Log On


   

Question 167728: Verify this identity :
(tan^2(x)-1)/(1+tan^2(x)) = 1-2cos^2(x)
I've started a couple different options but none are working out for me. Here is what I have so far :
A)
multiply top by (1 + tan^2(x)) to get :
tan^4 (x) -1 or (tan^2(x)+1)(tan^2x-1)
then i'm stuck!
B) (tanx +1)(tanx-1)/1 + tan^2(x) =
(sinx/cosx + 1)(sinx/cosx - 1) / 1/cosx
then again I'm stuck!
I'm not sure if I should be working on the right side of the equation instead! Grrrrr....I could get a little help from the tutors in the Math Lab on campus but we've been instructed not to seek them out for this problem (and another I am stuck on!). Any tips would be helpful and thank you in advance for your time!!
Desperately seeking solution, Rebecca
Found 5 solutions by jim_thompson5910, saravpreetminhas, Alan3354, MathTherapy, ikleyn:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Note: I'm only manipulating the left side


... Start with the given equation



... Use the identity to simplify the denominator


... Now use the identity to get the denominator in terms of cosine


... Multiply the first fraction by the reciprocal of the second fraction


... Get tangent in terms of sine and cosine


... Distribute


... Multiply and simplify


... Factor out a negative 1


... Rearrange the terms.



... Use the identity



... Use the identity


... Distribute


... Rearrange the terms.


Since the two sides are equal, this confirms the identity

Answer by saravpreetminhas(1) About Me  (Show Source):
You can put this solution on YOUR website!
1-2cos^2(x)={[sin^2(x)/cos^2(x)]-1}/{[sin^2(x)/cos^2(x)]+1}

1-2cos^2(x)= {[sin^2(x)-cos^2(x)]/[sin^2(x)+cos^2(x)]}
1-2cos^2(x) = sin^2(x)-cos^2(x)
1 = sin^2(x)-cos^2(x) + 2cos^2(x)
1 = sin^2(x)+cos^2(x)
1 = 1

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Verify this identity :
(tan^2(x)-1)/(1+tan^2(x)) = 1-2cos^2(x)
Verify this identity :
(tan^2(x)-1)/sec^2 = 1-2cos^2(x)
Mulitiply by cos^2
(sin^2 - cos^2)/1 = 1 - 2cos^2
Add cos^2
sin^2 = 1 - cos^2
QED

Answer by MathTherapy(10549) About Me  (Show Source):
You can put this solution on YOUR website!

Verify this identity :
(tan^2(x)-1)/(1+tan^2(x)) = 1-2cos^2(x)
I've started a couple different options but none are working out for me. Here is what I have so far :
A)
multiply top by (1 + tan^2(x)) to get :
tan^4 (x) -1 or (tan^2(x)+1)(tan^2x-1)
then i'm stuck!
B) (tanx +1)(tanx-1)/1 + tan^2(x) =
(sinx/cosx + 1)(sinx/cosx - 1) / 1/cosx
then again I'm stuck!
I'm not sure if I should be working on the right side of the equation instead! Grrrrr....I could get a little help from the tutors in the Math Lab on campus but we've been instructed not to seek them out for this problem (and another I am stuck on!). Any tips would be helpful and

thank you in advance for your time!!
Desperately seeking solution, Rebecca
It's much, much easier than you think

%28tan%5E2+%28x%29+-+1%29%2F%281+%2B+tan%5E2+%28x%29%29+=+1+-+2cos%5E2+%28x%29



Left side:

%28sec%5E2+%28x%29+-+1+-+1%29%2F%281+%2B+sec%5E2+%28x%29+-+1%29 ------ Replacing matrix%281%2C3%2C+tan%5E2+%28x%29%2C+with%2C+sec%5E2+%28x%29%29

%28sec%5E2+%28x%29+-+2%29%2Fsec%5E2+%28x%29

%281%2Fcos%5E2+%28x%29+-+2%29%2F%281%2Fcos%5E2+%28x%29%29 ------ Replacing matrix%281%2C3%2C+sec%5E2+%28x%29%2C+with%2C+1%2Fcos%5E2+%28x%29%29

1%2Fcos%5E2+%28x%29+-+2%22%F7%221%2Fcos%5E2+%28x%29 ------ Making above LOOK less complex

%281+-+2cos%5E2+%28x%29%29%2Fcos%5E2+%28x%29%22%F7%221%2Fcos%5E2+%28x%29 

%281+-+2cos%5E2+%28x%29%29%2Fcos%5E2+%28x%29%22%2A%22cos%5E2+%28x%29%2F1 ----- Applying KEEP, CHANGE, FLIP

%281+-+2cos%5E2+%28x%29%29%2Fcross%28cos%5E2+%28x%29%29%22%2A%22cross%28cos%5E2+%28x%29%29%2F1 ----- Canceling numerator and denominator

Left side:highlight_green%281+-+2cos%5E2+%28x%29%29, identical to Right side:highlight_green%281+-+2cos%5E2+%28x%29%29

Answer by ikleyn(52754) About Me  (Show Source):
You can put this solution on YOUR website!
.
Verify this identity :
(tan^2(x)-1)/(1+tan^2(x)) = 1-2cos^2(x)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

It is done in 3 (three) easy steps. 

1.  The numerator tan^2(x) - 1 = sin%5E2%28x%29%2Fcos%5E2%28x%29+-1 = %28sin%5E2%28x%29-cos%5E2%28x%29%29%2Fcos%5E2%28x%29.                 (1)


2.  The denominator 1 + tan^2(x) = 1+%2B+sin%5E2%28x%29%2Fcos%5E2%28x%29 = %28sin%5E2%28x%29+%2B+cos%5E2%28x%29%29%2Fcos%5E2%28x%29 = 1%2Fcos%5E2%28x%29.      (2)


3.  When you divide the numerator (expression (1)) by the denominator (expression (2)), cos^2(x) cancels, 
    and the remaining expression is  sin%5E2%28x%29-cos%5E2%28x%29,  which you can rewrite

    sin%5E2%28x%29-cos%5E2%28x%29 = %281-cos%5E2%28x%29%29+-+cos%5E2%28x%29 = 1+-+2%2Acos%5E2%28x%29.


    It is precisely your right side.


For other similar solved problems see the lesson
    - Proving Trigonometry identities
in this site.


Also, you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this online textbook under the topic "Trigonometry: Solved problems".


Do not be desperate !
Ask in this forum.