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- 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.

And use the relation of tan and cot i.e.. $\tan \theta .\cot \theta =1\ or\ \tan \theta =\dfrac{1}{\cot \theta }$.

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.

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