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

Solution

## The correct option is C $$3$$Range of $$\sec^{-1}x\space :\space [0,\pi/2)\cup(\pi/2,\pi]$$So $$\sec^{-1}x+\sec^{-1}y+\sec^{-1}z=3\pi$$ only when all of them are equal to $$\pi$$That is, $$\sec^{-1}x=\pi,\sec^{-1}y=\pi,\sec^{-1}z=\pi$$$$\sec^{-1}x=\pi \rightarrow x=-1$$Similarly, $$y=-1,z=-1$$$$\therefore \space xy+yz+zx=1+1+1=3$$Hence, $$(C)$$Mathematics

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