If $\tan A-\tan B=x\ and\ \cot B-\cot A=y,$ prove that $\cot \left( A-B \right)=\dfrac{1}{x}+\dfrac{1}{y}$.

- Hint:Here, we do not need to calculate exact values of trigonometric functions. Just use the given equations to get a relationship of equations for proving. Find the value of $\tan A-\tan B,\tan A.\tan B$ using the given equations, now put these values in the trigonometric identity $\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$
And use the relation of tan and cot i.e.. $\tan \theta .\cot \theta =1\ or\ \tan \theta =\dfrac{1}{\cot \theta }$.

Complete step-by-step solution -

To prove
$\cot \left( A-B \right)=\dfrac{1}{x}+\dfrac{1}{y}.................\left( i \right)$
It is given that $\tan A-\tan B=x\ and\ \cot B-\cot A=y$. Let us suppose them in form of equations as,
$\tan A-\tan B=x...........\left( ii \right)$
$\cot B-\cot A=y............\left( iii \right)$
Let us take the LHS part of equation (i) and simplify it to get RHS.
So, LHS = cot (A – B)
Now, we know $\tan \theta =\dfrac{1}{\cot \theta }$ , so $\cot \left( A-B \right)$ can be written in form of tan as,
$LHS=\dfrac{1}{\tan \left( A-B \right)}...........\left( iv \right)$
Now, we know the trigonometric identity of tan (A – B) as,
$\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}.............\left( v \right)$
Hence, LHS of equation (iv) can be written with the help of equation (v) as,
$LHS=\dfrac{1}{\dfrac{\tan A-\tan B}{1+\tan A\tan B}}=\dfrac{1+\tan A\tan B}{\tan A-\tan B}............\left( vi \right)$
Now, let us simplify equation (iii), we get,
$\cot B-\cot A=y$
We have $\cot \theta =\dfrac{1}{\tan \theta }$, so we get,
\[\begin{align}
  & \dfrac{1}{\tan B}-\dfrac{1}{\tan A}=y \\
 & \Rightarrow \dfrac{\tan A-\tan B}{\tan A\tan B}=y \\
\end{align}\]
From equation (ii), $\tan A-\tan B=x$, so we can rewrite the above equation as,
$\begin{align}
  & \dfrac{x}{\tan A\tan B}=y \\
 & \Rightarrow \tan A\tan B=\dfrac{x}{y}.............\left( vii \right) \\
\end{align}$
Now, putting the values of $\tan A.\tan B,\ \tan A-\tan B$ from equation (vii) and (ii) respectively in the equation (vi), we get
$\begin{align}
  & LHS=\dfrac{1+\left( \dfrac{x}{y} \right)}{x}=\dfrac{1}{x}+\left( \dfrac{x}{y} \right)\left( \dfrac{1}{x} \right) \\
 & \Rightarrow LHS=\dfrac{1}{x}+\dfrac{1}{y}=RHS \\
\end{align}$
So, LHS = RHS from the above equation. Hence, it is proved that,
$\cot \left( A-B \right)=\dfrac{1}{x}+\dfrac{1}{y}$

Note: Another approach for proving the given relation that we can put values of ‘x’ and ‘y’ in RHS and simplify it further to get LHS i.e. $cot\left( A-B \right)$.
Don’t confuse the formula of $\tan (A-B),\tan (A+B)$. One may use $\tan \left( A-B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ which is wrong. So, take care with the trigonometric identities to solve these kinds of questions.
No need for exact values of trigonometric functions. One can waste his/her time getting values from them. Just use the given relationships to prove the given equation.