Question

Prove that:

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

Answer 1

$({\sec }^{6}x-{\tan }^{6}x)$

$=({\sec }^{2}x{)}^3-({\tan }^{2}x{)}^3$

$=({\sec }^{2}x-{\tan }^{2}x{)}^3+3{\sec }^{2}x\cdot {\tan }^{2}x({\sec }^{2}x-{\tan }^{2}x)$

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

$[\because {\sec }^{2}x-{\tan }^{2}x=1]$.