Prove that $\dfrac{{\operatorname{Sin} A}}{{1 + \operatorname{Cos} A}} + \dfrac{{1 + \operatorname{Cos} A}}{{\operatorname{Sin} A}} = 2\operatorname{Cos} ecA$

Hint: Here first we will take the L.C.M of L.H.S equation.
We know that \[{\sin ^2}A + {\cos ^2}A = 1\] and $\dfrac{1}{{\sin A}} = \cos ecA$, simply solve the left hand of equation using these identities and get the value of RHS

Complete step-by-step answer:
Let’s start solving with left hand equation
We have $\dfrac{{\sin A}}{{1 + \cos A}} + \dfrac{{1 + \cos A}}{{\sin A}}$ ……….(1)
Taking L.C.M of (1)
\[
   = \dfrac{{\operatorname{Sin} A\left( {\operatorname{Sin} A} \right) + \left( {1 + \operatorname{Cos} A} \right)\left( {1 + \operatorname{Cos} A} \right)}}{{\left( {1 + \operatorname{Cos} A} \right)(\operatorname{Sin} A)}} \\
   = \dfrac{{{{\operatorname{Sin} }^2}A + {{(1 + \operatorname{Cos} A)}^2}}}{{\left( {1 + \operatorname{Cos} A} \right)(\operatorname{Sin} A)}} \\
 \]
Apply the formula of ${(a + b)^2} = {a^2} + 2ab + {b^2}$ for ${(1 + \cos A)^2}$ we get:-
\[
   = \dfrac{{{{\operatorname{Sin} }^2}A + {{\left( 1 \right)}^2} + 2\left( 1 \right)\left( {\operatorname{Cos} A} \right) + {{\left( {\operatorname{Cos} A} \right)}^2}}}{{\left( {1 + \operatorname{Cos} A} \right)(\operatorname{Sin} A)}} \\
   = \dfrac{{{{\operatorname{Sin} }^2}A + 1 + 2\operatorname{Cos} A + {{\operatorname{Cos} }^2}A}}{{(1 + \operatorname{Cos} A)(\operatorname{Sin} A)}} \\
 \]
We know that ${\operatorname{Sin} ^2}A + {\operatorname{Cos} ^2}A = 1$
Therefore putting in the value we get:-
\[
   = \dfrac{{1 + 1 + 2\operatorname{Cos} A}}{{(1 + \operatorname{Cos} A)(\operatorname{Sin} A)}}\; \\
   = \dfrac{{2 + 2\operatorname{Cos} A}}{{(1 + \operatorname{Cos} A)(\operatorname{Sin} A)}} \\
 \]
Take 2 as common from numerator we get:-
\[ = \dfrac{{2(1 + \operatorname{Cos} A)}}{{(1 + \operatorname{Cos} A)(\operatorname{Sin} A)}}\]
Cancelling the terms we get:-
\[ = \dfrac{2}{{\operatorname{Sin} A}}\]
\[ = 2.\dfrac{1}{{\operatorname{Sin} A}}\]
Now we know that,
\[\dfrac{1}{{\operatorname{Sin} A}} = \operatorname{Cos} ecA\]
Therefore putting in the value we get:-
\[ = 2\operatorname{Cos} ecA\]
\[ = {\text{ }}RHS\]
Since \[LHS = {\text{ }}RHS\]
Hence proved

Note:
In these types of questions students just need to use basic identities of trigonometry then just simplify your question and then put these identities to get your answer.
The identities used should be correct and accurate.
Also, Cosec is the reciprocal of sine