Prove that cos2x= 1-tan^2x/ 1+tan^2x
Thanks =)
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} $
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)