If the equation tan4x−2sec2x+a2=0 has at least one solution, then sum of all possible integral values of ‘a’ is equal to

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answer is A.

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Detailed Solution

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

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