Question 1109239
If you use {{{cos(A)/sin(A)}}} for cot(A); and {{{sin(A)/cos(A)}}} for tan(A), and do the algebraic steps, that will lead you to:


{{{(1/(cos(A)sin(A)))^2}}}


and then  {{{1/((cos(A))^2(sin(A))^2)}}}


Use of the basic pythagorean identity,
{{{1/((1-sin^2(A))(sin(A))^2)}}}
which is all in terms of sines.