# Question #dcdc4

Aug 1, 2017

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

Divide both sides of the equation by ${\cos}^{2} \left(\theta\right)$

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

Substitute ${\tan}^{2} \left(\theta\right)$ for ${\sin}^{2} \frac{\theta}{\cos} ^ 2 \left(\theta\right)$:

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

Substitute ${\sec}^{2} \left(\theta\right)$ for $\frac{1}{\cos} ^ 2 \left(\theta\right)$:

$1 + {\tan}^{2} \left(\theta\right) = {\sec}^{2} \left(\theta\right)$ Q.E.D.

Aug 1, 2017

See the proof below

#### Explanation:

We need

$\tan \theta = \sin \frac{\theta}{\cos} \theta$

$\sec \theta = \frac{1}{\cos} \theta$

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$
Dividing throughout by ${\cos}^{2} \theta$
${\sin}^{2} \frac{\theta}{\cos} ^ 2 \theta + {\cos}^{2} \frac{\theta}{\cos} ^ 2 \theta = \frac{1}{\cos} ^ 2 \theta$
${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$
$Q E D$