If f(x)=cot−1(2x1−x2) and g(x)=cos−1(1−x21+x2) , then limx→af(x)−f(a)g(x)−g(a)(0

f(x)−f(a)g(x)−g(a)={π2−tan−1(2x1−x2)}−{π2−tan−1(2a1−a2)}cos−1(1−x21+x2)−cos−1(1−a21+a2) =tan−1(2a1−a2)−tan−1(2x1−x2)cos−1(1−x21+x2)−cos−1(1−a21+a2)=tan−1(tan2α)−tan−1(tan2θ)cos−1(cos2θ)−cos−1(cos2α) , where      x=tanθ,a=tanα .=2(α−θ)2(θ−α)=−1