Prove the following statement:
\[\dfrac{\sin A}{1+\cos A}+\dfrac{1+\cos A}{\sin A}=2\operatorname{cosec}A\]

Hint: In this question, consider the LHS and simplify it by taking the LCM sin A (1 + cos A) and using \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. Now, take out the common term and after canceling the common terms, use \[\dfrac{1}{\sin \theta }=\operatorname{cosec}\theta \] to prove the desired result.
Complete step-by-step answer:
Here, we have to prove that \[\dfrac{\sin A}{1+\cos A}+\dfrac{1+\cos A}{\sin A}=2\operatorname{cosec}A\]
Let us consider the LHS of the given expression,
\[E=\dfrac{\sin A}{1+\cos A}+\dfrac{1+\cos A}{\sin A}\]
By taking (1 + cos A) sin A as LCM and simplifying the above expression, we get,
\[E=\dfrac{\left( \sin A \right).\left( \sin A \right)+\left( 1+\cos A \right)\left( 1+\cos A \right)}{\left( 1+\cos A \right).\sin A}\]
We can also write the above expression as,
\[E=\dfrac{{{\sin }^{2}}A+{{\left( 1+\cos A \right)}^{2}}}{\left( 1+\cos A \right)\sin A}\]
We know that, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]. By using it in the above expression, we get,
\[E=\dfrac{{{\sin }^{2}}A+{{\left( 1 \right)}^{2}}+\left( {{\cos }^{2}}A \right)+2\cos A}{\left( 1+\cos A \right)\sin A}\]
By rearranging the terms of the above expression, we get,
\[E=\dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A+1+2\cos A}{\left( 1+\cos A \right)\sin A}\]
We know that, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. By using this in the above expression, we get,
\[E=\dfrac{1+1+2\cos A}{\left( 1+\cos A \right)\sin A}\]
\[E=\dfrac{2+2\cos A}{\left( 1+\cos A \right)\sin A}\]
By taking 2 common in the numerator of the above expression, we get,
\[E=\dfrac{2\left( 1+\cos A \right)}{\left( 1+\cos A \right)\sin A}\]
By canceling (1 + cos A) from numerator and denominator of the above expression, we get,
\[E=\dfrac{2}{\sin A}\]
We know that \[\dfrac{1}{\sin \theta }=\operatorname{cosec}\theta \]. By using this in the above expression, we get,
E = 2 cosec A
E = RHS
So, LHS = RHS
Hence proved.
So, we have proved that
\[\dfrac{\sin A}{1+\cos A}+\dfrac{1+\cos A}{\sin A}=2\operatorname{cosec}A\]

Note: In this question, by looking at RHS of the question, that is 2 cosec A, some students substitute \[\cos A=\dfrac{1}{\sec A}\] and \[\sin A=\dfrac{1}{\operatorname{cosec}A}\] which is not needed and which only makes the solution lengthy and confusing. Students should also first simplify the given expression which is in terms of \[\sin \theta \] and \[\cos \theta \] and then take out the common terms and use formulas to prove the desired result.