Prove that: $\dfrac{\sin A-2{{\sin }^{3}}A}{2{{\cos }^{3}}A-\cos A}=\tan A$.

Hint: In order to solve this question, we will express $2{{\sin }^{3}}A=\left( \sin A2{{\sin }^{2}}A \right)$ and $2{{\cos }^{3}}A=\left( \cos A2{{\cos }^{2}}A \right)$ and we will make use of the formula, ${{\sin }^{2}}A+{{\cos }^{2}}A=1$. We will first write $\left( \sin A2{{\sin }^{2}}A \right)$ in place of $2{{\sin }^{3}}A$ and $\left( \cos A2{{\cos }^{2}}A \right)$ in place of $2{{\cos }^{3}}A$ and then we will solve the question further in order to prove the expression.

Complete step by step solution:
It is given in the question that we have to prove $\dfrac{\sin A-2{{\sin }^{3}}A}{2{{\cos }^{3}}A-\cos A}=\tan A$. We will start from the LHS first, so we can write the LHS as,
$\dfrac{\sin A-2{{\sin }^{3}}A}{2{{\cos }^{3}}A-\cos A}$
We know that $2{{\sin }^{3}}A$ can also be written as $\left( \sin A2{{\sin }^{2}}A \right)$ and $2{{\cos }^{3}}A$ can be written as $\left( \cos A2{{\cos }^{2}}A \right)$. So, on replacing $2{{\sin }^{3}}A$ by $\left( \sin A2{{\sin }^{2}}A \right)$ and $2{{\cos }^{3}}A$ by $\left( \cos A2{{\cos }^{2}}A \right)$ in the above expression, we will get,
\[\dfrac{\sin A-\sin A2{{\sin }^{2}}A}{\cos A2{{\cos }^{2}}A-\cos A}\]
On taking sin A common from the numerator, we get,
\[\dfrac{\sin A\left( 1-2{{\sin }^{2}}A \right)}{2{{\cos }^{3}}A-\cos A}\]
Now, on taking cos A common from the denominator, we get,
\[\dfrac{\sin A\left( 1-2{{\sin }^{2}}A \right)}{\cos A\left( 2{{\cos }^{2}}A-1 \right)}\]
Now, we know that ${{\sin }^{2}}A+{{\cos }^{2}}A=1$. So on replacing 1 by ${{\sin }^{2}}A+{{\cos }^{2}}A$, we get,
\[\dfrac{\sin A\left( {{\sin }^{2}}A+{{\cos }^{2}}A-2{{\sin }^{2}}A \right)}{\cos A\left( 2{{\cos }^{2}}A-{{\sin }^{2}}A-{{\cos }^{2}}A \right)}\]
\[\dfrac{\sin A\left( {{\cos }^{2}}A-{{\sin }^{2}}A \right)}{\cos A\left( {{\cos }^{2}}A-{{\sin }^{2}}A \right)}\]
Now, on cancelling the like terms in the numerator and the denominator, we get,
\[\dfrac{\sin A}{\cos A}\]
Now, we know that \[\dfrac{\sin A}{\cos A}\] is equal to tan A. Therefore, we get the LHS as,
LHS = tan A
We have the RHS as tan A. So, we can say,
LHS = RHS
Hence proved.

Note: The possible mistake that the students can make in this question is in the step where we have to substitute ${{\sin }^{2}}A+{{\cos }^{2}}A$ in place of 1 in the expression \[\dfrac{\sin A\left( 1-2{{\sin }^{2}}A \right)}{\cos A\left( 2{{\cos }^{2}}A-1 \right)}\]. Most of the students, after substituting the value get the expression as, \[\dfrac{\sin A\left( {{\sin }^{2}}A+{{\cos }^{2}}A-2{{\sin }^{2}}A \right)}{\cos A\left( 2{{\cos }^{2}}A-{{\sin }^{2}}A+{{\cos }^{2}}A \right)}\] but this is wrong, as in the denominator of \[\dfrac{\sin A\left( 1-2{{\sin }^{2}}A \right)}{\cos A\left( 2{{\cos }^{2}}A-1 \right)}\] there is a negative sign before 1, hence when we substitute ${{\sin }^{2}}A+{{\cos }^{2}}A$, the negative sign must be applied to it also, and we should get,
\[\dfrac{\sin A\left( {{\sin }^{2}}A+{{\cos }^{2}}A-2{{\sin }^{2}}A \right)}{\cos A\left( 2{{\cos }^{2}}A-{{\sin }^{2}}A-{{\cos }^{2}}A \right)}\]. So, the students must be careful while solving this question.