Question

If ${\sec }^{-1}x+{\sec }^{-1}y+{\sec }^{-1}z=3\pi$, then $xy+yz+zx=$ _______.

• A. $0$
• B. $-3$
• C. $3$
• D. $1$

Answer 1

The correct option is C $3$

Range of ${\sec }^{-1}x:[0,\pi /2)\cup (\pi /2,\pi ]$

So ${\sec }^{-1}x+{\sec }^{-1}y+{\sec }^{-1}z=3\pi$ only when all of them are equal to $\pi$

That is, ${\sec }^{-1}x=\pi ,{\sec }^{-1}y=\pi ,{\sec }^{-1}z=\pi$

${\sec }^{-1}x=\pi \to x=-1$

Similarly, $y=-1,z=-1$

$\therefore xy+yz+zx=1+1+1=3$

Hence, $\left(C\right)$