Prove that the trigonometric equation: ${\cot ^2}A - {\cos ^2}A = {\cos ^2}A.{\cot ^2}A$

Hint: To prove this identity first we have to start from the LHS side and on transforming one trigonometric function into another we have to reach RHS. We have to use some standard results to prove this question.

Complete Step-by-Step solution:
We have on LHS side
\[{\cot ^2}A - {\cos ^2}A\]
We can write it as
 $\dfrac{{{{\cos }^2}A}}{{{{\sin }^2}A}} - {\cos ^2}A{\text{ }}\left( {\because {{\cot }^2}A = \dfrac{{{{\cos }^2}A}}{{{{\sin }^2}A}}} \right)$
On taking common we get,
$
   \Rightarrow {\cos ^2}A\left( {\dfrac{1}{{{{\sin }^2}A}} - 1} \right) = {\cos ^2}A\left( {\dfrac{{1 - {{\sin }^2}A}}{{{{\sin }^2}A}}} \right) \\
   \Rightarrow {\cos ^2}A.\dfrac{{{{\cos }^2}A}}{{{{\sin }^2}A}}{\text{ }}\left( {\because 1 - {{\sin }^2}A = {{\cos }^2}A} \right) \\
$
$ \Rightarrow {\cos ^2}A.{\cot ^2}A$ =RHS
Hence proved.

Note: Whenever we get this type of question the key concept of solving is you have to start either from LHS or RHS side whichever is easy and proceed using your basic knowledge to get required terms. Here you should notice that if any formulae can’t be applied then break the terms then you can apply the formula.