How do you simplify tan^4 theta + 2tan^2 theta + 1 ?

Oct 18, 2015

${\tan}^{4} \left(\theta\right) + 2 {\tan}^{2} \left(\theta\right) + 1 = {\sec}^{4} \left(\theta\right)$

Explanation:

${\tan}^{4} \left(\theta\right) + 2 {\tan}^{2} \left(\theta\right) + 1$ is a perfect square, if you can remember the formula ${\left(x + y\right)}^{2} = {x}^{2} + 2 x y + {y}^{2}$

So we can say

${\tan}^{4} \left(\theta\right) + 2 {\tan}^{2} \left(\theta\right) + 1 = {\left({\tan}^{2} \left(\theta\right) + 1\right)}^{2}$

However, as we know, from the pythagorean identity

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

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

So,

${\tan}^{4} \left(\theta\right) + 2 {\tan}^{2} \left(\theta\right) + 1 = {\left({\sec}^{2} \left(\theta\right)\right)}^{2}$

${\tan}^{4} \left(\theta\right) + 2 {\tan}^{2} \left(\theta\right) + 1 = {\sec}^{4} \left(\theta\right)$