The equation y=sinxsin(x+2)−sin2(x+1) represents a straight line lying in :
A
first, second and fourth quadrants
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B
second and third quadrants only
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C
third and fourth quadrants only
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D
first, third and fourth quadrants
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Solution
The correct option is C third and fourth quadrants only y=12[2sinx⋅sin(x+2)−2sin2(x+1)] ⇒y=12[cos2−cos(2x+2)−1+cos(2x+2)] ⇒y=12(cos2−1)=−sin21<0
Hence the straight line lying in third and fourth quadrants only.