# If tan A- tan B=x and cot B- cot A=y , then cot ( A-B)=?

A) $\dfrac{1}{y}-\dfrac{1}{x}$

B) $\dfrac{1}{x}-\dfrac{1}{y}$

C) $\dfrac{1}{x}+\dfrac{1}{y}$

D) None of these

Answer

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Hint: The given problem is related to trigonometric formulae for compound angles. Try to remember the formulae related to $\tan $ and $\cot $ of compound angles.

Complete step-by-step answer:

To solve the given problem, the following formulae will be used:

1) $\cot \left( A-B \right)=\dfrac{\cot A\cot B+1}{\cot B-\cot A}$

2) $\tan A\cot A=1$

In the question, we are asked to find the value of $\cot \left( A-B \right)$ in terms of $x$ and $y$ , where the value of \[x\] is given as $x=\tan A-\tan B$ and the value of $y$ is given as $y=\cot B-\cot A$.

Now, we know, $\cot \left( A-B \right)=\dfrac{\cot A\cot B+1}{\cot B-\cot A}$.

We will write the RHS as two fractions separated by an additional sign.

So, we get $\cot \left( A-B \right)=\dfrac{\cot A\cot B}{\cot B-\cot A}+\dfrac{1}{\cot B-\cot A}$.

Now, we know, the value of $cotB-\cot A$ is given as $y$. So, we can write $\dfrac{1}{cotB-\cot A}$ as $\dfrac{1}{y}$.

So, $\cot \left( A-B \right)=\dfrac{\cot A\cot B}{\cot B-\cot A}+\dfrac{1}{y}$.

Now, we know, we can write $\dfrac{\cot A\cot B}{\cot B-\cot A}$ as $\dfrac{1}{\dfrac{\cot B-\cot A}{\cot A\cot B}}$, which can further be written as $\dfrac{1}{\dfrac{1}{\cot A}-\dfrac{1}{\cot B}}$.

So, $\cot \left( A-B \right)=\dfrac{1}{\dfrac{1}{\cot A}-\dfrac{1}{\cot B}}+\dfrac{1}{y}$.

Now, we know, $\tan A\cot A=1$. So, $\dfrac{1}{\cot A}=\tan A$ and $\dfrac{1}{\cot B}=\tan B$.

So, $\cot \left( A-B \right)=\dfrac{1}{\tan A-\tan B}+\dfrac{1}{y}$.

Now, we are given that $\tan A-\tan B=x$. So, $\dfrac{1}{\tan A-\tan B}=\dfrac{1}{x}$.

So, $\cot \left( A-B \right)=\dfrac{1}{x}+\dfrac{1}{y}$.

Hence, $\cot \left( A-B \right)$ can be written as $\dfrac{1}{x}+\dfrac{1}{y}$ , where the value of \[x\] is given as $x=\tan A-\tan B$ and the value of $y$ is given as $y=\cot B-\cot A$.

Therefore, the correct option is option C.

Note: Most of the students get confused in the formula of $\cot \left( A-B \right)$. The value of $\cot \left( A-B \right)$ is given as $\cot \left( A-B \right)=\dfrac{\cot A\cot B+1}{\cot B-\cot A}$. But generally, students get confused and write the value of $\cot \left( A-B \right)$ as $\cot \left( A-B \right)=\dfrac{\cot A\cot B+1}{\cot A-\cot B}$, which is wrong. Such confusion should be avoided. This confusion can lead to wrong answers. Also, while making substitutions of $x$ and $y$ , be careful of the sign convention. Sign mistakes are common but can lead to wrong answers and hence, should be avoided

Complete step-by-step answer:

To solve the given problem, the following formulae will be used:

1) $\cot \left( A-B \right)=\dfrac{\cot A\cot B+1}{\cot B-\cot A}$

2) $\tan A\cot A=1$

In the question, we are asked to find the value of $\cot \left( A-B \right)$ in terms of $x$ and $y$ , where the value of \[x\] is given as $x=\tan A-\tan B$ and the value of $y$ is given as $y=\cot B-\cot A$.

Now, we know, $\cot \left( A-B \right)=\dfrac{\cot A\cot B+1}{\cot B-\cot A}$.

We will write the RHS as two fractions separated by an additional sign.

So, we get $\cot \left( A-B \right)=\dfrac{\cot A\cot B}{\cot B-\cot A}+\dfrac{1}{\cot B-\cot A}$.

Now, we know, the value of $cotB-\cot A$ is given as $y$. So, we can write $\dfrac{1}{cotB-\cot A}$ as $\dfrac{1}{y}$.

So, $\cot \left( A-B \right)=\dfrac{\cot A\cot B}{\cot B-\cot A}+\dfrac{1}{y}$.

Now, we know, we can write $\dfrac{\cot A\cot B}{\cot B-\cot A}$ as $\dfrac{1}{\dfrac{\cot B-\cot A}{\cot A\cot B}}$, which can further be written as $\dfrac{1}{\dfrac{1}{\cot A}-\dfrac{1}{\cot B}}$.

So, $\cot \left( A-B \right)=\dfrac{1}{\dfrac{1}{\cot A}-\dfrac{1}{\cot B}}+\dfrac{1}{y}$.

Now, we know, $\tan A\cot A=1$. So, $\dfrac{1}{\cot A}=\tan A$ and $\dfrac{1}{\cot B}=\tan B$.

So, $\cot \left( A-B \right)=\dfrac{1}{\tan A-\tan B}+\dfrac{1}{y}$.

Now, we are given that $\tan A-\tan B=x$. So, $\dfrac{1}{\tan A-\tan B}=\dfrac{1}{x}$.

So, $\cot \left( A-B \right)=\dfrac{1}{x}+\dfrac{1}{y}$.

Hence, $\cot \left( A-B \right)$ can be written as $\dfrac{1}{x}+\dfrac{1}{y}$ , where the value of \[x\] is given as $x=\tan A-\tan B$ and the value of $y$ is given as $y=\cot B-\cot A$.

Therefore, the correct option is option C.

Note: Most of the students get confused in the formula of $\cot \left( A-B \right)$. The value of $\cot \left( A-B \right)$ is given as $\cot \left( A-B \right)=\dfrac{\cot A\cot B+1}{\cot B-\cot A}$. But generally, students get confused and write the value of $\cot \left( A-B \right)$ as $\cot \left( A-B \right)=\dfrac{\cot A\cot B+1}{\cot A-\cot B}$, which is wrong. Such confusion should be avoided. This confusion can lead to wrong answers. Also, while making substitutions of $x$ and $y$ , be careful of the sign convention. Sign mistakes are common but can lead to wrong answers and hence, should be avoided

Last updated date: 28th Sep 2023

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