Question #f837c

May 11, 2017

I tried this:

Explanation:

We can use the fact that:
$\sec \theta = \frac{1}{\cos} \theta$
and
$\tan \theta = \frac{\sin \theta}{\cos \theta}$

and write:

$\left(1 - \cos \theta\right) \left(1 + \cos \theta\right) \frac{1}{\cancel{{\cos}^{2} \theta}} = {\sin}^{2} \frac{\theta}{\cancel{{\cos}^{2} \theta}}$

multiply on the left to get:

$1 - {\cos}^{2} \theta = {\sin}^{2} \theta$ that it is true!

Remebering that:

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

May 15, 2017

$\left(1 - \cos \theta\right) \left(1 + \cos \theta\right) {\sec}^{2} \theta = \left(1 - {\cos}^{2} \theta\right) \cdot \frac{1}{\cos} ^ 2 \theta$

note : ${\sec}^{2} \theta = \frac{1}{\cos} ^ 2 \theta$ and $1 - {\cos}^{2} \theta = {\sin}^{2} \theta$

$\left(1 - {\cos}^{2} \theta\right) \cdot \frac{1}{\cos} ^ 2 \theta = {\sin}^{2} \frac{\theta}{\cos} ^ 2 \theta = {\tan}^{2} \theta$ --> proved