The equation y=sinx sin⁡(x+2)- sin^2⁡(x+1) represents a straight line lying in:

$$y=\mathrm{sin}⁡(x+2)\mathrm{sin}⁡x-{\mathrm{sin}}^{2}⁡(x+1)$$y=sin⁡(x+2)sin⁡x−sin2⁡(x+1)

$$=\frac{1}{2}[2\mathrm{sin}⁡(x+2)\mathrm{sin}⁡x-2{\mathrm{sin}}^{2}⁡(x+1)]$$=21​[2sin⁡(x+2)sin⁡x−2sin2⁡(x+1)]

[\because 2\mathrm{sin}⁡A\mathrm{sin}⁡B =\mathrm{cos}⁡(A-B)-\mathrm{cos}⁡(A+B)&\mathrm{cos}⁡2A=1-2{\mathrm{sin}}^{2}⁡A]

$$=\frac{1}{2}[\mathrm{cos}⁡(x+2-x)-\mathrm{cos}⁡(2x+2)-(1-\mathrm{cos}⁡2(x+1))]$$=21​[cos⁡(x+2−x)−cos⁡(2x+2)−(1−cos⁡2(x+1))]

$$=\frac{1}{2}[\mathrm{cos}⁡2-\mathrm{cos}⁡(2x+2)-1+\mathrm{cos}⁡(2x+2)]$$=21​[cos⁡2−cos⁡(2x+2)−1+cos⁡(2x+2)]

$$=\frac{1}{2}[1-\mathrm{cos}⁡2]=-\frac{1}{2}2{\mathrm{sin}}^{2}⁡1$$=21​[1−cos⁡2]=−21​2sin2⁡1

$=-{\mathrm{sin}}^{2}⁡1 ($=−sin2⁡1( constant and negative $)$)

Hence, the straight line lies in the third and fourth quadrants only.