Quotient rule states if #g(x)=(a(x))/(b(x))#
then #(dg)/(dx)=((da)/(dx)xxb(x)-(db)/(dx)xxa(x))/(b(x))^2#
Here #a(x)=cotx+cscx# and #b(x)=tanx-sinx#
i.e. #(da)/(dx)=-csc^2x-cotxcscx# and #(db)/(dx)=sec^2x-cosx#
Hence #(dg(x))/(dx)=([(tanx-sinx)(-csc^2x-cotxcscx)-(sec^2x-cosx)(cotx+cscx)])/(tanx-sinx)^2#
= #([-tanxcsc^2x-cscx+cscx+cotx
-secxcscx-sec^2xcscx+cosxcotx+cotx])/(tanx-sinx)^2#
= #([-tanxcsc^2x+2cotx
-secxcscx-sec^2xcscx+cosxcotx])/(tanx-sinx)^2#
We can simplify it further by converting each trigonometric ratio into #sinx# and #cosx# and simplifying it further
but the simpler way could be that as #g(x)=(cotx+cscx)/(tanx-sinx)#
= #(cosx/sinx+1/sinx)/(sinx/cosx-sinx)#
= #((cosx+1)/sinx)/((sinx-sinxcosx)/(sinxcosx)#
= #(cosx(1+cosx))/(sinx(1-cosx))#
=#cotxcot^2(x/2)#
and hence #(dg)/(dx)=2cotxcot(x/2)(-cot(x/2)csc(x/2))xx1/2+cot^2(x/2)(-cotxcscx)#
= #-cotxcot^2(x/2)csc(x/2)-cotxcscxcot^2(x/2)#
= #-cotxcot^2(x/2)(csc(x/2)+cscx)#
