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Question: If \[\tan \alpha - \tan \beta = m\] and \[\cot \alpha - \cot \beta = n\], then prove that \(\cot \le...

If tan⁡α−tan⁡β=m\tan \alpha - \tan \beta = m and cot⁡α−cot⁡β=n\cot \alpha - \cot \beta = n, then prove that cot⁡(α−β)=1m−1n\cot \left( {\alpha - \beta } \right) = \dfrac{1}{m} - \dfrac{1}{n}.

Explanation

Solution

This type of problem should always start with taking the left-hand side or by taking the right-hand side. Here we will take the left-hand side and then expand the equation by using the formula cot⁡(α−β)=cot⁡α.cot⁡β+1cot⁡β−cot⁡α\cot \left( {\alpha - \beta } \right) = \dfrac{{\cot \alpha .\cot \beta + 1}}{{\cot \beta - \cot \alpha }}. So by using this we will prove this question.

Formula used:
cot⁡(α−β)=cot⁡α.cot⁡β+1cot⁡β−cot⁡α\cot \left( {\alpha - \beta } \right) = \dfrac{{\cot \alpha .\cot \beta + 1}}{{\cot \beta - \cot \alpha }}

Complete step by step solution:
we have the equation cot⁡(α−β)=1m−1n\cot \left( {\alpha - \beta } \right) = \dfrac{1}{m} - \dfrac{1}{n}
Now taking the left-hand side, we get
⇒cot⁡(α−β)\Rightarrow \cot \left( {\alpha - \beta } \right)
Now by using the formula we can write the above equation as
⇒cot⁡(α−β)=cot⁡α.cot⁡β+1cot⁡β−cot⁡α\Rightarrow \cot \left( {\alpha - \beta } \right) = \dfrac{{\cot \alpha .\cot \beta + 1}}{{\cot \beta - \cot \alpha }}
And since it is given that cot⁡α−cot⁡β=n\cot \alpha - \cot \beta = n
Therefore, cot⁡β−cot⁡α=−n\cot \beta - \cot \alpha = - n
So putting it in the equation, we get
⇒cot⁡(α−β)=cot⁡α.cot⁡β+1−n\Rightarrow \cot \left( {\alpha - \beta } \right) = \dfrac{{\cot \alpha .\cot \beta + 1}}{{ - n}}, and we will name it equation 11
Now since tan⁡α−tan⁡β=m\tan \alpha - \tan \beta = m
So it can also be written as
⇒1cot⁡α−1cot⁡β=m\Rightarrow \dfrac{1}{{\cot \alpha }} - \dfrac{1}{{\cot \beta }} = m
Now taking the LCM, we can write it as
⇒cot⁡β−cot⁡αcot⁡α.cot⁡β=m\Rightarrow \dfrac{{\cot \beta - \cot \alpha }}{{\cot \alpha .\cot \beta }} = m
And as we know, cot⁡β−cot⁡α=−n\cot \beta - \cot \alpha = - n
Therefore, the above equation can be written as
⇒−ncot⁡α.cot⁡β=m\Rightarrow \dfrac{{ - n}}{{\cot \alpha .\cot \beta }} = m
And also it can be written as
⇒cot⁡α.cot⁡β=−nm\Rightarrow \cot \alpha .\cot \beta = \dfrac{{ - n}}{m}, and we will name it equation 22
Now on substituting the value in the equation11, we get
⇒cot⁡α.cot⁡β+1−n=−nm+1−n\Rightarrow \dfrac{{\cot \alpha .\cot \beta + 1}}{{ - n}} = \dfrac{{\dfrac{{ - n}}{m} + 1}}{{ - n}}
And on solving the above equation, we can write it as
⇒−n+m−nm\Rightarrow \dfrac{{ - n + m}}{{ - nm}}
And it can also be written as
⇒−n−nm+m−nm\Rightarrow \dfrac{{ - n}}{{ - nm}} + \dfrac{m}{{ - nm}}
On canceling out the same term, we get
⇒1m−1n\Rightarrow \dfrac{1}{m} - \dfrac{1}{n}
Hence, the above equation is proved.

Therefore cot⁡(α−β)=1m−1n\cot \left( {\alpha - \beta } \right) = \dfrac{1}{m} - \dfrac{1}{n}.

Note:
There is one other way to prove this question by using the formula called tan⁡(α−β)=tan⁡α−tan⁡βtan⁡α.tan⁡β+1\tan \left( {\alpha - \beta } \right) = \dfrac{{\tan \alpha - \tan \beta }}{{\tan \alpha .\tan \beta + 1}} and from this we can make the equation cot⁡(α−β)\cot \left( {\alpha - \beta } \right) by interchanging the numerator by the denominator. And then on further solving we will come to the conclusion which will state that LHS will be equal to RHS. So this type of problem has various ways to prove and we have to decide which process is easy. I would prefer the first method as it is easier and convenient.