If tan−1(x−4x−2)+tan−1(x+4x+2)=4π, find the value of x.Answer: x=±2SolutionCombine with the tangent-sum formula:tan−1[1−x−4x−2⋅x+4x+2x−4x−2+x+4x+2]=4πNumerator: (x−2)(x+4)+(x+2)(x−4)=(x2+2x−8)+(x2−2x−8)=2x2−16.Denominator: (x−4)(x+4)−(x−2)(x+2)=(x2−16)−(x2−4)=−12.−122x2−16=tan4π=1 ⇒ 2x2−16=−12 ⇒ 2x2=4 ⇒ x2=2x=±2CBSE Class 12 Board · 2014 · Subjective