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Q.

The locus of the point  cosec θ+cot θ,  cosec θ-cot θ  is

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a

xy+1=0

b

x2-y2=1

c

xy=1

d

x2+y2=1

answer is C.

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Detailed Solution

 Let P=(x, y) be locus of point.

Given that Given (x, y)=(cosecθ+cotθ, cosecθ-cotθ)  x= cosecθ+cotθ   -   (1)       y= cosecθ+cotθ   -   (2) (1)×(2) xy=(cosecθ+cotθ) (cosecθ-cotθ)               xy=cosec2θ-cot2θ              xy=1

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