Prove the identity $\left( {\sec A - \sin A} \right)\left( {\cos {\text{ec}}A + \cos A} \right) = {\sin ^2}A\tan A + \cot A$

Hint: We will prove the identity by simplifying the left-hand-side and right-hand-side of the given expression using the trigonometric formulas. Simplify the LHS by opening the brackets and using the identities like $\sec A = \dfrac{1}{{\cos A}}$ and $\operatorname{cosec} A = \dfrac{1}{{\sin A}}$ and convert the given expression in terms of sine and cosine. Similarly, use properties $\tan A = \dfrac{{\sin A}}{{\cos A}}$ and $\cot A = \dfrac{{\cos A}}{{\sin A}}$ to simplify RHS.

Complete step-by-step answer:
We have to prove the identity $\left( {\sec A - \sin A} \right)\left( {\cos {\text{ec}}A + \cos A} \right) = {\sin ^2}A\tan A + \cot A$
We will begin by solving L.H.S
Multiply the brackets on L.H.S to simplify the expression.
$\sec A\operatorname{cosec} A + \sec A\cos A - \sin A\operatorname{cosec} A - \sin A\cos A$
Now, we now that $\sec A = \dfrac{1}{{\cos A}}$ and $\operatorname{cosec} A = \dfrac{1}{{\sin A}}$
Therefore, we have,
$
  \dfrac{1}{{\cos A}}\left( {\dfrac{1}{{\sin A}}} \right) + \left( {\dfrac{1}{{\cos A}}} \right)\cos A - \sin A\left( {\dfrac{1}{{\sin A}}} \right) - \sin A\cos A \\
   \Rightarrow \dfrac{1}{{\cos A}}\left( {\dfrac{1}{{\sin A}}} \right) + 1 - 1 - \sin A\cos A \\
   \Rightarrow \dfrac{1}{{\cos A}}\left( {\dfrac{1}{{\sin A}}} \right) - \sin A\cos A \\
$
On simplifying the above expression, we will get,
$\dfrac{{1 - {{\sin }^2}A{{\cos }^2}A}}{{\cos A\sin A}}$
But, we know that ${\sin ^2}A + {\cos ^2}A = 1$
Then,
$\dfrac{{{{\sin }^2}A + {{\cos }^2}A - {{\sin }^2}A{{\cos }^2}A}}{{\cos A\sin A}}$
We will take \[{\sin ^2}A\] common from the numerator and simplify it further,
\[\dfrac{{{{\cos }^2}A + {{\sin }^2}A\left( {1 - {{\cos }^2}A} \right)}}{{\cos A\sin A}}\]
Also, $1 - {\cos ^2}A = {\sin ^2}A$
$
  \dfrac{{{{\cos }^2}A + {{\sin }^2}A\left( {{{\sin }^2}A} \right)}}{{\cos A\sin A}} \\
   \Rightarrow \dfrac{{{{\cos }^2}A + {{\sin }^4}A}}{{\cos A\sin A}} \\
$
Now, we will simplify R.H.S of the given expression,
${\sin ^2}A\tan A + \cot A$
We know that $\tan A = \dfrac{{\sin A}}{{\cos A}}$ and $\cot A = \dfrac{{\cos A}}{{\sin A}}$
Therefore, the expression in the R.H.S can be written as,
$
  {\sin ^2}A\left( {\dfrac{{\sin A}}{{\cos A}}} \right) + \dfrac{{\cos A}}{{\sin A}} \\
   \Rightarrow \dfrac{{{{\sin }^3}A}}{{\cos A}} + \dfrac{{\cos A}}{{\sin A}} \\
$
Now, we will take the L.C.M
$\dfrac{{{{\sin }^4}A + {{\cos }^2}A}}{{\cos A\sin A}}$
Which is equal to LHS.
Since, LHS is equal to RHS, the given identity is proved.

Note: Students must remember the trigonometric identities to do these types of questions. Simplification plays an important role in these questions. We can also start from LHS and simplify it form an expression given in RHS.