Prove that ${{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A$.

Hint: In this question, we have to prove ${{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$ to be equal to ${{\tan }^{2}}A$. For this, we will first change tanA and cotA in terms of sinA and cosA and then simplify the expression to get ${{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$ to be equal to ${{\tan }^{2}}A$.
We will use the trigonometric formula to evaluate our answer.
\[\begin{align}
  & \left( i \right)\tan \theta =\dfrac{\sin \theta }{\cos \theta } \\
 & \left( ii \right)\cot \theta =\dfrac{\cos \theta }{\sin \theta } \\
\end{align}\]

Complete step-by-step answer:
Here we are given the equation as, ${{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A$.
Let us solve the left side of the equation to make it equal to the right side of the equation.
Left side of the equation is given as ${{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$.
We know that, we can separate the squares for numerator and denominator so we get: $\dfrac{{{\left( 1-\tan A \right)}^{2}}}{{{\left( 1-\cot A \right)}^{2}}}$.
As we know that, $\tan \theta $ is equal to $\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta $ is equal to $\dfrac{\cos \theta }{\sin \theta }$. So putting in these values we get:
$\dfrac{{{\left( 1-\dfrac{\sin A}{\cos A} \right)}^{2}}}{{{\left( 1-\dfrac{\cos A}{\sin A} \right)}^{2}}}$.
Now let us take LCM of cosA in the numerator and LCM of sinA in the denominator, we get:
$\dfrac{{{\left( \dfrac{\cos A-\sin A}{\cos A} \right)}^{2}}}{{{\left( \dfrac{\sin A-\cos A}{\sin A} \right)}^{2}}}$.
Simplifying the denominator we get:
${{\left( \dfrac{\cos A-\sin A}{\cos A} \right)}^{2}}/{{\left( \dfrac{\sin A-\cos A}{\sin A} \right)}^{2}}$.
Combining the squares we get:
${{\left( \dfrac{\cos A-\sin A}{\cos A}\times \dfrac{\sin A}{\sin A-\cos A} \right)}^{2}}$.
Taking negative sign common from cosA-sinA we get:
${{\left( \dfrac{-\left( \sin A-\cos A \right)\sin A}{\cos A\left( \sin A-\cos A \right)} \right)}^{2}}$.
Cancelling (sinA-cosA) from the numerator and the denominator we get: ${{\left( \dfrac{-\sin A}{\cos A} \right)}^{2}}$.
We know that, $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ so we can write above expression as ${{\left( -\tan A \right)}^{2}}$.
As we know that, square of any negative term becomes positive so we get: ${{\left( \tan A \right)}^{2}}$ which can be written as ${{\tan }^{2}}A$.
This is the right side of the equation.
Hence proved.

Note: Students should take care of the signs while cancelling terms. Take LCM in the numerator and the denominator carefully. Students should note that, ${{\tan }^{2}}\theta \text{ and }{{\left( \tan \theta \right)}^{2}}$ are same. Students should keep in mind all the basic properties of trigonometry before solving these sums.