Prove the following equation:
\[\left( {{\text{cosec }}A - \sin A} \right)\left( {{\text{sec }}A - \cos A} \right) = \dfrac{1}{{\tan A + \cot A}}\]

Hint: In order to prove this question, we will use the basic formulas of trigonometric function which will help us to convert sec, cosec, sin and cosine terms in the form of tan and cot. We will use the following formulas
${\text{cosec}}\theta = \dfrac{1}{{\sin \theta }},\sec \theta = \dfrac{1}{{\cos \theta }},{\sin ^2}\theta + {\cos ^2}\theta = 1$

Complete step-by-step answer:
Given equation is
\[\left( {{\text{cosec }}A - \sin A} \right)\left( {{\text{sec }}A - \cos A} \right) = \dfrac{1}{{\tan A + \cot A}}\]
Taking LHS
\[ = \left( {{\text{cosec }}A - \sin A} \right)\left( {{\text{sec }}A - \cos A} \right)\]
Using the formulas $\left[ {{\text{cosec}}\theta = \dfrac{1}{{\sin \theta }},\sec \theta = \dfrac{1}{{\cos \theta }}} \right]$
$
   = \left( {\dfrac{1}{{\sin A}} - \sin A} \right)\left( {\dfrac{1}{{\cos A}} - \cos A} \right) \\
   = \dfrac{{1 - {{\sin }^2}A}}{{\sin A}}\left( {\dfrac{{1 - {{\cos }^2}A}}{{\cos A}}} \right) \\
 $
Simplifying it further
\[
   = \dfrac{{{{\cos }^2}A}}{{\sin A}}\left( {\dfrac{{{{\sin }^2}A}}{{\cos A}}} \right) \\
   = \dfrac{{\cos A \times \sin A}}{1} \\
   = \dfrac{{\cos A \times \sin A}}{{{{\cos }^2}A + {{\sin }^2}A}}{\text{ As we know that }}\left[ {{{\cos }^2}\theta + {{\sin }^2}\theta = 1} \right] \\
   = \dfrac{1}{{\dfrac{{{{\cos }^2}A + {{\sin }^2}A}}{{\cos A \times \sin A}}}} \\
   = \dfrac{1}{{\dfrac{{\cos A}}{{\sin A}} + \dfrac{{\sin A}}{{\cos A}}}} \\
   = \dfrac{1}{{\tan A + \cot A}} \\
 \]
Hence LHS = RHS
Hence proved.

Note: In order to solve such problems related to trigonometric simplification and proof. Students must remember some basic trigonometric identities for a fast solution. Also any substitutions which are made should be done keeping in mind the RHS of the problem.