1. ## [SOLVED] Trigonometry Help!?

Prove the identity $\frac{1-tan^2x}{1+tan^2x} = 1-2sin^2x$

2. Hello,

$\tan(x)=\frac{\sin(x)}{\cos(x)}$

$\frac{1-tan^2x}{1+tan^2x} = \frac{1-\frac{\sin^2x}{\cos^2x}}{1+\frac{\sin^2x}{\cos^2x} }$ $= \frac{\frac{\cos^2x-\sin^2x}{\cos^2x}}{\frac{\cos^2x+\sin^2x}{cos^2x}}$ $= \frac{\cos^2x-\sin^2x}{\underbrace{\cos^2x+\sin^2x}_{1}}$ $= \underbrace{\cos^2x}_{1-\sin^2x}-\sin^2x = 1-2 \sin^2x$

3. Hello, looi76!

A slightly different approach . . .

$\frac{1-\tan^2\!x}{1+\tan^2\!x} = 1-2\sin^2\!x$
We have: . $\frac{1-\tan^2\!x}{\sec^2\!x} \;=\;\frac{1 - \dfrac{\sin^2\!x}{\cos^2\!x}}{\dfrac{1}{\cos^2\!x} }$

Multiply by $\frac{\cos^2\!x}{\cos^2\!x}\!:\quad\frac{\cos^2\!x \left(1 - \dfrac{\sin^2\!x}{\cos^2\!x}\right)} {\cos^2\!x\left(\dfrac{1}{\cos^2\!x}\right)} \;=\;\frac{\cos^2\!x - \sin^2\!x}{1}$

Then: . $\cos^2\!x - \sin^2\!x \;\;=\;\;(1-\sin^2\!x) - \sin^2\!x\;\;=\;\;1-2\sin^2\!x$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Hello, moo!

Beautiful job with the LaTeX!

Take a look at how I use \dfrac to make enlarged fractions.
(I learned about it only a few months ago.)

4. Hi Soroban !

Actually, i've learnt the use of \underbrace while reading one of your messages

$\sum_{k=0}^\infty \dfrac{x^n \ \text{which \ is \ the \ nth \ power \ of \ x}}{n!}$

$\sum_{k=0}^\infty \frac{x^n \ \text{which \ is \ the \ nth \ power \ of \ x}}{n!}$

erm... can't see the difference :s

$\frac{1-\tan^2\!x}{\sec^2\!x} \;=\;\frac{1 - \dfrac{\sin^2\!x}{\cos^2\!x}}{\dfrac{1}{\cos^2\!x} }$

$\frac{1-\tan^2\!x}{\sec^2\!x} \;=\;\frac{1 - \frac{\sin^2\!x}{\cos^2\!x}}{\frac{1}{\cos^2\!x}}$

Ow, i see :-)

Thanks !

5. what is sec ?

6. $sec(x)$ is just another name for $\frac{1}{cos(x)}$

7. And to explain :

$\sec^2x=\frac{1}{\cos^2x}=\dfrac{\cos^2x+\sin^2x}{ \cos^2x}=1+\dfrac{\sin^2x}{cos^2x}$ $=1+\tan^2x$