# How do you simplify 3(tan^2 theta - sec^2 theta)?

Aug 5, 2015

$3 \left({\tan}^{2} \left(\theta\right) - {\sec}^{2} \left(\theta\right)\right) = - 3$

#### Explanation:

$\tan \left(\theta\right) = \frac{\sin \left(\theta\right)}{\cos \left(\theta\right)}$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$\rightarrow {\tan}^{2} \left(\theta\right) = {\sin}^{2} \frac{\theta}{\cos} ^ 2 \left(\theta\right)$

$\sec \left(\theta\right) = \frac{1}{\cos} \left(\theta\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$${\sec}^{2} \left(\theta\right) = \frac{1}{\cos} ^ 2 \left(\theta\right)$

${\tan}^{2} \left(\theta\right) - {\sec}^{2} \left(\theta\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$= {\sin}^{2} \frac{\theta}{\cos} ^ 2 \left(\theta\right) - \frac{1}{\cos} ^ 2 \left(\theta\right)$

$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$=(sin^2(theta)-1)/(cos^2(theta)

and since ${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$= \frac{- {\cos}^{2} \left(\theta\right)}{\cos} ^ 2 \left(\theta\right)$

$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$= - 1$

$3 \left({\tan}^{2} \left(\theta\right) - {\sec}^{2} \left(\theta\right)\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$= - 3$

Aug 6, 2015

You can use the identity:

$1 + {\tan}^{2} \theta = {\sec}^{2} \theta$

$\implies 3 \left({\tan}^{2} \theta - \left(1 + {\tan}^{2} \theta\right)\right)$

$= 3 \left({\tan}^{2} \theta - 1 - {\tan}^{2} \theta\right)$

$= 3 \left(- 1\right) = \textcolor{b l u e}{- 3}$