Given tan−1(2x)+tan−1(3x)=4π
⇒tan−1(1−2x×3x2x+3x)=4π
Applying tangent on both sides,
⇒1−6x25x=1[∵tan4π=1]
⇒5x=1−6x2
⇒6x2+5x−1=0
⇒6x2+6x−x−1=0
⇒(6x−1)(x+1)=0
⇒x=−1,61
But, since x≥0.
Hence, x=61
∴A has only one element and thus, it is a singleton.