Question

Prove that ${\tan }^4\varphi +{\tan }^2\varphi ={\sec }^4\varphi -{\sec }^2\varphi$, for $0^{\circ }\le \varphi \le {90}^{\circ }$

Answer 1

${\tan }^4\varphi +{\tan }^2\varphi$

$={\tan }^2\varphi ({\tan }^2\varphi +1)$

$={\tan }^2\varphi \left({\sec }^2\varphi \right)$

$=({\sec }^2\varphi -1){\sec }^2\varphi$

$={\sec }^4\varphi -{\sec }^2\varphi$