If the equation tan4x−2sec2x+a2=0 has at least one solution, then sum of all possible integral values of ‘a’ is equal to
0
4
2
3
tan4x−2tan2x+1+a2=0
⇒tan4x-2tan2x+1=3-a2
⇒tan2x−12=3−a2≥0
⇒a2≤3⇒−3≤a≤3
⇒Integral values of ‘a’ are -1, 0, 1