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


A
0
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B
3
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C
3
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D
1
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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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