If the solution set of inequality (cot-1x) (tan-1x)+(2-π2)cot-1x-3tan-1 x-3(2-π2)>0 is (a,b), then the value of cot-1 a+cot-1b is
(cot-1x) (tan-1x)+(2-π2)cot-1x-3tan-1 x-3(2-π2)>0 ⇒ cot-1x (tan-1x+2-π2)-3(tan-1 x+2-π2)>0 ⇒ (cot-1x-3) (tan-1x+2-π2)>0 Since tan-1x-π2 =-cot-1x, ⇒(cot-1x-3)(2-cot-1x) >0 ⇒(cot-1x-3)(cot-1x-2) <0 ⇒ 2<cot-1x<3 ⇒cot 3<x<cot 2 since it is decreasing functionHence x∈(cot3,cot2) cot-1 a+cot-1b= cot-1 (cot3)+cot-1(cot2) =3+2 =5