The given expression is sec2(x+y)−1+cos2(x+y)+y2+2y=0 sec2(x+y)+cos2(x+y)+y2+2y+1=2 For the above expression to be true x+y=0 and y2+2y+1=0 x=−y and (y+1)2=0⇒y=−1 Hence point P and Q coincides and so d=0. cosd=1 x+y=π and y2+2y+1=0 x=π+1,y=−1 So, cosd=−1 Therefore, cosd=(−1)n,n∈N