Proving Trig. Identity

**MordernWar2** · January 20th 2010, 07:34 PM

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

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

OR

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

Please, try you best.
Thank you.

**pickslides** · January 20th 2010, 08:04 PM

Originally Posted by MordernWar2

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

I get.

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

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

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

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

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

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

$<br /> \csc(x)<br />$

**Soroban** · January 20th 2010, 08:05 PM

Hello, MordernWar2!

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

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

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

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

$2)\;\;\sin x +\sin x\cot^2\!x \:=\: \sec x$

This is not an identity . . . I suspect a typo.

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

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

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

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