Question

Solve

${\sec }^{6}x-{\tan }^{6}x-3{\sec }^{2}x.{\tan }^2$

Answer 1

We know that $a^3-b^3=(a-b)(a^2+ab+b^2)$

Now, ${\sec }^{6}x-{\tan }^{6}x=({\sec }^{2}x-{\tan }^{2}x)({\sec }^{4}x+{\sec }^{2}x{\tan }^{2}x+{\tan }^{4}x)=1.({\sec }^{4}x+{\sec }^{2}x{\tan }^{2}x+{\tan }^{4}x)=({\sec }^{2}x-{\tan }^{2}x{)}^2+2{\sec }^{2}x{\tan }^{2}x+{\tan }^{2}x{\sec }^{2}x=1+3{\sec }^{2}x{\tan }^{2}x$

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