Concept:Use the sum formula for inverse tangent: tan−1A+tan−1B=tan−1(1−ABA+B). When the sum equals 2π, the argument inside tan−1 becomes infinite, so the denominator 1−AB must be zero.Explanation:Given equation: tan−1(1+x)+tan−1(1−x)=2π.Apply the formula: tan−1(1−(1+x)(1−x)(1+x)+(1−x))=2π.Simplify numerator: (1+x)+(1−x)=2.Simplify denominator: 1−(1+x)(1−x)=1−(1−x2)=x2.So we get: tan−1(x22)=2π.For tan−1(value)=2π, the value must be infinite (since tan2π is undefined). Hence x22→∞, which forces x2=0.Thus x=0.Answer:x=0 (Option C).