For the equation 1−2x−x2=tan2x+y+cot2x+y

1−2x−x2=tan2x+y+cot2x+y⇒1−2x−x2-2=tan2x+y+cot2x+y-2⇒−x+12=tanx+y−cotx+y2Now L.Η.S≤0 and R.Η.δ≥0⇒−x+12=tanx+y−cotx+y2=0⇒            x=−1 and tanx+y=cotx+y ⇒           x=−1 and tan2−1+y=1   ⇒ x=−1 and −1+y=nπ±π4:n∈z⇒y=1+nπ±π4