How to prove this trigonometric identity?

[kasiviv002]

Nevermind, i figured it out, dont know hoe to delete this thread though :s

 

[Deleted member 4993]

(1-tan²x)/(1+tan²x) = Cos²x - sin²x

i'm not sure how to begin :s

convert:

\(\displaystyle tan(x) = \dfrac{sin(x)}{cos(x)}\)

 

[soroban]

Hello, kasiviv002!

\(\displaystyle \text{Prove that: }\:\dfrac{1-\tan^2\!x}{1+\tan^2\!x} \:=\:\cos^2\!x - \sin^2\!x\)

\(\displaystyle \dfrac{1-\tan^2\!x}{1+\tan^2\!x} \;=\;\dfrac{1-\tan^2\!x}{\sec^2\!x} \;=\;\cos^2\!x(1-\tan^2\!x)\)

. . . . . . . . .\(\displaystyle =\;\cos^2\!x\left(1 - \frac{\sin^2\!x}{\cos^2\!x}\right) \;=\;\cos^2\!x - \sin^2\!x\)