Can someone explain how this simplifies to one. Im very confused.

[eternal cape]

help

[fallow zealot]

Identities:

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

$\ \frac{d}{dx}tan(x) = sec(x) = \frac{1}{cos(x)} \$

Knowing this we can rewrite:

$\ sec^2(\theta) - tan^2(\theta) = \bigg(\frac{1}{cos(\theta)}\bigg)^2 - \bigg(\frac{sin(\theta)}{cos(\theta)}\bigg)^2\$

$\ = \frac{1}{cos^2(\theta)} - \frac{sin^2(\theta)}{cos^2(\theta)} = \frac{1-sin^2(\theta)}{cos^2(\theta)}\$

Now we also know:

$\ sin^2(x) + cos^2(x) = 1 \Rightarrow 1 -sin^2(x) = cos^2(x)\$

So it follows:

$\ = \frac{1-sin^2(\theta)}{cos^2(\theta)} = \frac{cos^2(\theta)}{cos^2(\theta)} = 1\$

[frosty panther]

adonhs

[eternal cape]

thank you bro, makes sense now

[fallow zealot]

.solved