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

Thanks =)

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- May 6th 2010, 01:37 PMHelloooidentity
Prove that cos2x= 1-tan^2x/ 1+tan^2x

Thanks =) - May 6th 2010, 02:09 PMharish21
- May 6th 2010, 02:24 PMSoroban
Hello, Hellooo!

Going from left to right is tricky . . .

Quote:

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} $

- May 6th 2010, 03:37 PMHellooo
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 >.< - May 6th 2010, 05:02 PMsa-ri-ga-ma
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) - May 7th 2010, 03:03 PMHellooo
oh thnx a lot!! =))