1. ## Proving Trig Equations

Need help with these two, by tonight, if possible!

1) {sin^2x - tanx}/{cos^2x - cotx} = tan^2x

In words, sin squared x minus tan x all divided by cos squared x minus cotx equals tan squared x.

2) {sin^3x + cos^3x}/{1 - 2cos^2x} = {secx - sinx}/{tanx - 1}

In words, sin cubed x minus cos cubed x all divided by one minus two cos squared x equals sec x minus sin x all divided by tanx minus one.

2. Rewrite #1 and make the cancellations.

$\frac{sin^{2}x-tanx}{cos^{2}x-cotx}$

$sin^{2}x-tanx=tanx(sinxcosx-1)$

$cos^{2}x-cotx=\frac{sinxcosx-1}{tanx}$

3. Originally Posted by Blue Griffin
s tan squared x.

2) {sin^3x + cos^3x}/{1 - 2cos^2x} = {secx - sinx}/{tanx - 1}

In words, sin cubed x minus cos cubed x all divided by one minus two cos squared x equals sec x minus sin x all divided by tanx minus one.
Hmmmm....

$\frac{sin^3(x) + cos^3(x)}{1 - 2cos^2(x)} = \frac{(sin(x) + cos(x))(sin^2(x) - sin(x)cos(x) + cos^2(x))}{1 - 2cos^2(x)}$

$\frac{sin^3(x) + cos^3(x)}{1 - 2cos^2(x)} = \frac{(sin(x) + cos(x))(1 - sin(x)cos(x))}{1 - 2cos^2(x)}$

Divide both the numerator and denominator by cos^2(x). In the numerator divide each factor by cos(x):
$= \frac{(tan(x) + 1)(sec(x) - sin(x))}{sec^2(x) - 2}$

Now, $sec^2(x) - 1 = tan^2(x)$, so
$= \frac{(tan(x) + 1)(sec(x) - sin(x))}{tan^2(x) - 1}$

$= \frac{(tan(x) + 1)(sec(x) - sin(x))}{(tan(x) + 1)(tan(x) - 1)}$

$= \frac{sec(x) - sin(x)}{tan(x) - 1}$

-Dan

(Thank you. I enjoyed doing that one! )