# How do you verify the identity 3sec^2thetatan^2theta+1=sec^6theta-tan^6theta?

Oct 3, 2016

See below

#### Explanation:

$3 {\sec}^{2} \theta {\tan}^{2} \theta + 1 = {\sec}^{6} \theta - {\tan}^{6} \theta$

Right Side$= {\sec}^{6} \theta - {\tan}^{6} \theta$
$= {\left({\sec}^{2} \theta\right)}^{3} - {\left({\tan}^{2} \theta\right)}^{3}$->use difference of two cubes formula

=(sec^2theta-tan^2theta) (sec^4theta+sec^2thetatan^2theta+tan^4theta)

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

$= {\sec}^{4} \theta + {\sec}^{2} \theta {\tan}^{2} \theta + {\tan}^{4} \theta$

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

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

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

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

$= 3 {\sec}^{2} \theta {\tan}^{2} \theta + 1$

$=$ Left Side