
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
511.5k+ views
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
Recently Updated Pages
What percentage of the area in India is covered by class 10 social science CBSE

The area of a 6m wide road outside a garden in all class 10 maths CBSE

What is the electric flux through a cube of side 1 class 10 physics CBSE

If one root of x2 x k 0 maybe the square of the other class 10 maths CBSE

The radius and height of a cylinder are in the ratio class 10 maths CBSE

An almirah is sold for 5400 Rs after allowing a discount class 10 maths CBSE

Trending doubts
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Why is there a time difference of about 5 hours between class 10 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

Write an application to the principal requesting five class 10 english CBSE
