Question

If $P={\sec }^{6}A-{\tan }^{6}A$ and $Q=3{\sec }^{2}A{\tan }^{2}A,$ then $P-Q$ is equal to

• A. $0$
• B. $1$
• C. $2$
• D. $-1$

Answer 1

The correct option is B $1$

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