Prove the given trigonometric identity $\dfrac{{\sin A}}{{1 - \cos A}} = cosec A + \cot A$.

Hint: In L.H.S we are given one combination of trigonometric functions and just have to transform them. Try multiplying and dividing the L.H.S with the conjugate of denominator and apply basic math identity.

Complete step-by-step answer:
L.H.S
$\dfrac{{\sin A}}{{1 - \cos A}}$
Multiply and divide by $(1 + \cos A)$
$\dfrac{{\sin A}}{{1 - \cos A}} \times \dfrac{{1 + \cos A}}{{1 + \cos A}} = \dfrac{{\sin A(1 + \cos A)}}{{1 - {{\cos }^2}A}}$
[using identity $(a + b)(a - b) = {a^2} - {b^2}$]
$
   = \dfrac{{\sin A(1 + \cos A)}}{{{{\sin }^2}A}}{\text{ [}}\because {\text{si}}{{\text{n}}^2}A + {\cos ^2}A = 1] \\
   = \dfrac{{1 + \cos A}}{{\sin A}} \\
   = \dfrac{1}{{\sin A}} + \dfrac{{\cos A}}{{\sin A}} \\
   = { \ cosec A } + \cot A{\text{ [}}\because \sin A = \dfrac{1}{{ \ cosec A }}{\text{ and }}\cot A = \dfrac{{\cos A}}{{\sin A}}] \\
$
 L.H.S = R.H.S
Hence proved.

Note: In such types of trigonometric problems what we have to do is to simply make L.H.S equal to R.H.S or vice-versa and one should also know the basic conversion of trigonometric ratios and should also have the knowledge of trigonometric identities.