If #cot x=(2n^2+2mn)/(m^2+2mn)# then prove that
#csc^2x=1+((2n^2+2mn)/(m^2+2mn))^2# or
#csc^2x=((m^2+2mn)^2+(2n^2+2mn)^2)/(m^2+2mn)^2#
= #((m^4+4m^2n^2+4m^3n)+(4n^4+4m^2n^2+8mn^3))/(m^2+2mn)^2#
(using identity #(a+b)^2=a^2+b^2+2ab)#) and now adding and rearranging terms
#csc^2x=((m^4+4m^3n+8m^2n^2+8mn^3+4n^4))/(m^2+2mn)^2# ...(A)
Now #(m^2+2mn+2n^2)^2# becomes
#(m^4+4m^2n^2+4n^4+2*m^2*2mn+2*m^2*2n^2+2*2mn*2n^2)#
(using identity #(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc#)
or #(m^4+4m^2n^2+4n^4+4m^3n+4m^2n^2+8mn^3)#
or #(m^4+4m^3n+8m^2n^2+8mn^3+4n^4)# ....(B)
Note that in numerator of (A) and in (B), we have arranged monomials in decreasing degrees of #m# and increasing degree of #n#. Also note that both are equal.
Hence, #csc^2x=((m^2+2mn+2n^2)^2)/((m^2+2mn)^2)#
or #cscx=(m^2+2mn+2n^2)/(m^2+2mn)#
