Question

For any real number $x\ge 1$ the expression ${\sec }^2\left({\tan }^{-1}x\right)-{\tan }^2\left({\sec }^{-1}x\right)$ is equal to

• A. 1
• B. 2
• C. $2x^2$
• D. $2\sqrt{2}$

Answer 1

Answer: (B)

$1+{\tan }^2\left({\tan }^{-1}x\right)-\left({\sec }^2\left({\sec }^{-1}x\right)-1\right)$
$1+{\left(\tan \left({\tan }^{-1}x\right)\right)}^2-{\left(\sec \left({\sec }^{-1}x\right)\right)}^2+1=1+x^2-x^2+1=2$