f(x)=(32)cot3xcot2x;                              0

f(π−2)=limh→0(32)cot(3(π2−h))cot(2(π2−h)) =limh→0(32)tan3h−cot2h=limh→0(32)−(tan3h)(tan2h)=1f(π+2)=limh→0(1+|cot(π2+h)|)(a|tan(π2+h)|)b=limh→0(1+tanh)acothb=elimh→0(1+tanh−1)acothb=ea/bf(π2)=b+3f(x) is continuous atx=π/2⇒1=b+3=ea/b⇒b=−2  and a=0