Question

Prove that ${\sec }^{6}A+{\tan }^{6}A=1+3{\tan }^{2}A{\sec }^{2}A$

Answer 1

${\sec }^{6}A+{\tan }^{6}A=1+{\tan }^{2}A{\sec }^{2}ALHS={\sec }^{6}A+{\tan }^{6}A={({\sec }^{2}A-{\tan }^{2}A)}^3+{\tan }^{2}A{\sec }^{2}A({\sec }^{2}A-{\tan }^{2}A)=1+3{\tan }^{2}A{\sec }^{2}A$