The solution of tan−1⁡2x+tan−1⁡3x=π4 is

tan−1⁡2x+tan−1⁡3x=π4 ⇒ 3x+2x1−6x2=tan⁡π4⇒ 5x=1−6x2⇒ 6x2+5x−1=0⇒ x=−1,16But when x = -1,tan−1⁡2x=tan−1⁡(−2)<0 and  tan−1⁡3x=tan−1⁡(−3)<0This, value will not satisfy the given equation.Hence, x=16