Since sec2θ>1 ⇒[(n+1)sec2θ]>[nsec2θ] Hence, f and g are both one-one Let k<nsec2θ<k+1 and k<mcosec2θ<k+1 kcos2θ<n<(k+1)cos2θ and k1sin2θ<m<(k+1)sin2θ Adding gives k< integer <k+1 ⇒A∩B=φ Suppose k∉A ⇒nsec2θ<k and (k+1)<(n+1)sec2θ ⇒n<kcos2θ and n+1>(k+1)cos2θ ⇒k<(k−n)cosec2θ,k+1>(k−n)cosec2θ ⇒[(k−n)cosec2θ]=k ⇒k∈B