1. ## identity

Prove that cos2x= 1-tan^2x/ 1+tan^2x

Thanks =)

2. Originally Posted by Hellooo
Prove that cos2x= 1-tan^2x/ 1+tan^2x

Thanks =)
$\displaystyle RHS = \frac{1-tan^{2}x}{1+tan^{2}x} = \frac{1 - \frac{sin^{2}x}{cos^{2}x}}{1+\frac{sin^{2}x}{cos^{ 2}x}}$

try finishing it now

3. Hello, Hellooo!

Going from left to right is tricky . . .

Prove that: .$\displaystyle \cos2x\:=\:\frac{1-\tan^2\!x}{1+\tan^2\!x}$

We have: .$\displaystyle \cos2x \;=\;\frac{\cos^2\!x - \sin^2\!x}{1}$

Divide top and bottom by $\displaystyle \cos^2\!x\!:\quad \frac{\dfrac{\cos^2\!x}{\cos^2\!x} - \dfrac{\sin^2\!x}{\cos^2\!x}}{\dfrac{1}{\cos^2\!x} }$

. . and we have: .$\displaystyle \frac{1-\tan^2\!x}{\sec^2\!x} \;=\;\frac{1-\tan^2\!x}{1+\tan^2\!x}$

4. Thnx for the quick replies!!

if i didn't know that sec^2x =1+ tan^2x, is it a lot harder to prove that 1-tan^2x/1+tan^2x = cos2x?

i just did this in a test, and couldn't figure it out because i didn't memorize the sec, cot, or csc identities >.<

5. Originally Posted by Hellooo
Thnx for the quick replies!!

if i didn't know that sec^2x =1+ tan^2x, is it a lot harder to prove that 1-tan^2x/1+tan^2x = cos2x?
Not at all.
Take
(1-tan^2x)/(1+tan^2x)
Multiply and divide by cos^2x. You get
[cos^2x - tan^2x*cos^2x]/[cos^2x + tan^2x*cos^2x]
I hope, you know that tanx = sinx/cosx and basic indentity cos^2x + sin^2x = 1
So you get
[cos^2x - sin^2x]/[cos^2x + sin^2x]
= [cos^2x - sin^2x] = cos(2x)

6. oh thnx a lot!! =))