1. ## Proving Trig. Identity

Hey Guyzzz,
I have two Identities given and I have to prove either one.

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

OR

2) $\displaystyle sin x +sin x cot^2 x = sec x$

Thank you.

2. Originally Posted by MordernWar2

2) $\displaystyle sin x +sin x cot^2 x = sec x$
I get.

$\displaystyle \sin (x) +\sin (x) \cot^2 (x)$

$\displaystyle \sin (x) +\sin (x) \frac{\cos^2 (x)}{\sin^2(x)}$

$\displaystyle \sin (x) + \frac{\cos^2 (x)}{\sin(x)}$

$\displaystyle \frac{\sin^2 (x)}{\sin(x)} + \frac{\cos^2 (x)}{\sin(x)}$

$\displaystyle \frac{\sin^2 (x) + \cos^2 (x)}{\sin(x)}$

$\displaystyle \frac{1}{\sin(x)}$

$\displaystyle \csc(x)$

3. Hello, MordernWar2!

$\displaystyle 1)\;\;\frac{1 - \tan^2\!x}{1 + \tan^2\!x} \:=\: \cos 2x$

We have: .$\displaystyle \frac{1-\dfrac{\sin^2\!x}{\cos^2\!x}}{1 + \dfrac{\sin^2\!x}{\cos^2\!x}}$

Multiply by $\displaystyle \frac{\cos^2\!x}{\cos^2\!x}; \quad \frac{\cos^2\!x\left(1 - \dfrac{\sin^2\!x}{\cos^2\!x}\right)}{\cos^2\!x\lef t(1 + \dfrac{\sin^2\!x}{\cos^2\!x}\right)}$

. . . . . . . . $\displaystyle =\;\;\frac{\overbrace{\cos^2\!x - \sin^2\!x}^{\text{This is }\cos2x}}{\underbrace{\cos^2\!x + \sin^2\!x}_{\text{This is 1}}} \;\;=\;\; \cos2x$

$\displaystyle 2)\;\;\sin x +\sin x\cot^2\!x \:=\: \sec x$
This is not an identity . . . I suspect a typo.

It is probably: .$\displaystyle \sin x + \sin x\cot^2\!x \:=\: {\color{blue}\csc x}$

$\displaystyle \text{Factor the left side: }\:\sin x\underbrace{(1 + \cot^2\!x)}_{\text{This is }\csc^2x} \;\;=\;\;\sin x\csc^2\!x$

. . . . . . . . . . . . $\displaystyle =\;\;\underbrace{(\sin x\csc x)}_{\text{This is 1}}\csc x \;\;=\;\;csc x$