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
<pre>It's much, much easier than you think
{{{(tan^2 (x) - 1)/(1 + tan^2 (x)) = 1 - 2cos^2 (x)}}}

<b><u>Left side:</b></u>
{{{(sec^2 (x) - 1 - 1)/(1 + sec^2 (x) - 1)}}} ------ Replacing {{{matrix(1,3, tan^2 (x), with, sec^2 (x))}}}
{{{(sec^2 (x) - 2)/sec^2 (x)}}}
{{{(1/cos^2 (x) - 2)/(1/cos^2 (x))}}} ------ Replacing {{{matrix(1,3, sec^2 (x), with, 1/cos^2 (x))}}}
{{{1/cos^2 (x) - 2}}}{{{"÷"}}}{{{1/cos^2 (x)}}} ------ Making above LOOK less complex
{{{(1 - 2cos^2 (x))/cos^2 (x)}}}{{{"÷"}}}{{{1/cos^2 (x)}}} 
{{{(1 - 2cos^2 (x))/cos^2 (x)}}}{{{"*"}}}{{{cos^2 (x)/1}}} ----- Applying KEEP, CHANGE, FLIP
{{{(1 - 2cos^2 (x))/cross(cos^2 (x))}}}{{{"*"}}}{{{cross(cos^2 (x))/1}}} ----- Canceling numerator and denominator
<b><u>Left side:</b></u>{{{highlight_green(1 - 2cos^2 (x))}}}, identical to <b><u>Right side:</b></u>{{{highlight_green(1 - 2cos^2 (x))}}}