How do you prove $\dfrac{{\cos \,A}}{{1 - \tan A}} = \dfrac{{\sin A}}{{1 - \cot A}}$?

Hint: Here we have to solve the trigonometric functions. We have to prove left side of the equation to the right side of the equation. Firstly, we will solve left side of the equation by converting \[\tan \theta \] in terms of $\sin \theta $ and $\cos \theta $ by using the formula $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and then we will use different trigonometric identities to solve the equation such as $\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta = 2{\cos ^2}\theta - 1$ and similarly we will solve right side of the equation by converting $\cot \theta $ in terms of $\sin \theta $ and $\cos \theta $ by using the formula $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ and then we will use different trigonometric identities.

Complete step by step answer:
 We have $\dfrac{{\cos \,A}}{{1 - \tan A}} = \dfrac{{\sin A}}{{1 - \cot A}}$
Consider left side of the equation i.e., $\dfrac{{\cos \,A}}{{1 - \tan A}}$
We know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$. So,
$ \Rightarrow \dfrac{{\cos \,A}}{{1 - \tan A}} = \dfrac{{\cos \,A}}{{1 - \dfrac{{\sin A}}{{\cos A}}}}$
$ \Rightarrow \dfrac{{\cos \,A}}{{\dfrac{{\cos A - \sin A}}{{\cos A}}}}$
Simplifying the above equation. We get,
$ \Rightarrow \dfrac{{{{\cos }^2}\,A}}{{\cos A - \sin A}}$

Multiply and divide the above equation by the conjugate of $\cos A - \sin A$ i.e., $\cos A + \sin A$
$ \Rightarrow \dfrac{{{{\cos }^2}\,A}}{{\cos A - \sin A}} \times \dfrac{{\cos A + \sin A}}{{\cos A + \sin A}}$
Now using the identity $(a - b)(a + b) = {a^2} - {b^2}$. We get,
$ \Rightarrow \dfrac{{{{\cos }^2}\,A \times (\cos A + \sin A)}}{{{{\cos }^2}A - {{\sin }^2}A}}$
Now substituting ${\cos ^2}A - {\sin ^2}A = \cos 2A$. We get,
$ \Rightarrow \dfrac{{{{\cos }^2}\,A \times (\cos A + \sin A)}}{{\cos 2A}}$
Multiply and divide the above equation by $2$. We get,
$ \Rightarrow \dfrac{{2{{\cos }^2}\,A \times (\cos A + \sin A)}}{{2\cos 2A}}$

Now substituting $2{\cos ^2}A = \cos 2A - 1$ using the identity $\cos 2\theta = 2{\cos ^2}\theta - 1$. We get,
$ \Rightarrow \dfrac{{(\cos 2A - 1)(\cos A + \sin A)}}{{2\cos 2A}}$
Therefore, $\dfrac{{\cos \,A}}{{1 - \tan A}}$$ = \dfrac{{(\cos 2A - 1)(\cos A + \sin A)}}{{2\cos 2A}}$
Now, consider right side of the equation i.e., $\dfrac{{\sin A}}{{1 - \cot A}}$
We know that $\,\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$. So,
$ \Rightarrow \dfrac{{\sin \,A}}{{1 - \cot A}} = \dfrac{{\sin \,A}}{{1 - \dfrac{{\cos A}}{{\sin A}}}}$
$ \Rightarrow \dfrac{{\sin \,A}}{{\dfrac{{\sin A - \cos A}}{{\sin A}}}}$
Simplifying the above equation. We get,
$ \Rightarrow \dfrac{{{{\sin }^2}\,A}}{{\sin A - \cos A}}$

Taking negative signs common from the denominator. We get,
$ \Rightarrow \dfrac{{ - {{\sin }^2}\,A}}{{\cos A - \sin A}}$
Multiply and divide the above equation by the conjugate of $\cos A - \sin A$ i.e., $\cos A + \sin A$
$ \Rightarrow \dfrac{{ - {{\sin }^2}\,A}}{{\cos A - \sin A}} \times \dfrac{{\cos A + \sin A}}{{\cos A + \sin A}}$
Now using the identity $(a - b)(a + b) = {a^2} - {b^2}$. We get,
$ \Rightarrow \dfrac{{ - {{\sin }^2}\,A(\cos A + \sin A)}}{{{{\cos }^2}A - {{\sin }^2}A}}$
Now substituting ${\cos ^2}A - {\sin ^2}A = \cos 2A$. We get,
$ \Rightarrow \dfrac{{ - {{\sin }^2}\,A \times (\cos A + \sin A)}}{{\cos 2A}}$

Multiply and divide the above equation by $2$. We get,
$ \Rightarrow \dfrac{{ - 2{{\sin }^2}\,A \times (\cos A + \sin A)}}{{2\cos 2A}}$
Now substituting $2{\sin ^2}A = 1 - \cos 2A$ using the identity$\cos 2\theta = 1 - 2{\sin ^2}\theta $. We get
$ \Rightarrow \dfrac{{ - (1 - \cos 2A) \times (\cos A + \sin A)}}{{2\cos 2A}}$
Simplifying the above equation. We get,
$ \Rightarrow \dfrac{{(\cos 2A - 1) \times (\cos A + \sin A)}}{{2\cos 2A}}$
Therefore, $\dfrac{{\sin A}}{{1 - \cot A}}$$ = \dfrac{{(\cos 2A - 1)(\cos A + \sin A)}}{{2\cos 2A}}$

Since the left side of the equation is equal to the right side of the equation.Hence proved.

Note: In these types of problems in which we have to prove left side of the equation is equal to the right side of the equation if we are unable to prove one side equal to the other side we can solve both side of the equation and prove their results equal to one another. In these types of problems use trigonometric identity carefully and carefully identify by what number or equations we can multiply the equation. Note that if we are multiplying something in the numerator we also multiply in the denominator.