QUESTION

If ( tanA - tanB ) = x and ( cotB - cotA ) = y, then cot ( A - B ) isA.$\dfrac{1}{{x - y}}$ B.$\dfrac{1}{{x + y}}$ C.$\dfrac{1}{x} + y$ D.$\dfrac{1}{x} - \dfrac{1}{y}$ E.$\dfrac{1}{x} + \dfrac{1}{y}$

Hint-In this particular type of question we have to proceed by using the formula of$\cot \left( {A - B} \right) = \dfrac{{\cot B.\cot A + 1}}{{\cot B - \cot A}}$ and then rearranging the given values of tanA - tanB to put in the earlier equation. Then we need to simplify and get the desired value of cot ( A – B ).
$\begin{gathered} {\text{We know that }}\cot \left( {A - B} \right) = \dfrac{{\cot B.\cot A+1}}{{\cot B - \cot A}} \\ {\text{According to the question}} \\ \dfrac{1}{{\cot A}} - \dfrac{1}{{\cot B}} = x \\ \Rightarrow \dfrac{{\cot B - \cot A}}{{\cot A.cotB}} = x \\ \Rightarrow \cot B - \cot A = x\left( {\cot A.\cot B} \right) \\ \Rightarrow y = x\left( {\cot A.\cot B} \right){\text{ }}\left( {{\text{since }}\cot B - \cot A = y} \right) \\ \Rightarrow \dfrac{y}{x} = \left( {\cot A.\cot B} \right) \\ \end{gathered}$
Putting the values in $\cot \left( {A - B} \right) = \dfrac{{\cot B.\cot A + 1}}{{\cot B - \cot A}}$
$\cot \left( {A - B} \right) = \dfrac{{\dfrac{y}{x} + 1}}{y} = \dfrac{{\dfrac{{y + x}}{x}}}{y} = \dfrac{{y + x}}{{xy}} = \dfrac{1}{x} + \dfrac{1}{y}$
Note-Remember to recall the basic formulas of trigonometric functions while solving these types of questions. Note that $\tan \theta = \dfrac{1}{{\cot \theta }}$ . Also we need to understand the basic concept and rearrangement that is done while solving this question.