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Question: If ( tanA - tanB ) = x and ( cotB - cotA ) = y, then cot ( A - B ) is A.\(\dfrac{1}{{x - y}}\) ...

If ( tanA - tanB ) = x and ( cotB - cotA ) = y, then cot ( A - B ) is

A.1x−y\dfrac{1}{{x - y}}

B.1x+y\dfrac{1}{{x + y}}

C.1x+y\dfrac{1}{x} + y

D.1x−1y\dfrac{1}{x} - \dfrac{1}{y}

E.1x+1y\dfrac{1}{x} + \dfrac{1}{y}

Explanation

Solution

Hint-In this particular type of question we have to proceed by using the formula ofcot⁡(A−B)=cot⁡B.cot⁡A+1cot⁡B−cot⁡A\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 ).

Complete step-by-step answer:
( tanA - tanB ) = x and ( cotB - cotA ) = y
We know that cot⁡(A−B)=cot⁡B.cot⁡A+1cot⁡B−cot⁡A According to the question 1cot⁡A−1cot⁡B=x ⇒cot⁡B−cot⁡Acot⁡A.cotB=x ⇒cot⁡B−cot⁡A=x(cot⁡A.cot⁡B) ⇒y=x(cot⁡A.cot⁡B) (since cot⁡B−cot⁡A=y) ⇒yx=(cot⁡A.cot⁡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⁡(A−B)=cot⁡B.cot⁡A+1cot⁡B−cot⁡A\cot \left( {A - B} \right) = \dfrac{{\cot B.\cot A + 1}}{{\cot B - \cot A}}
We get,
cot⁡(A−B)=yx+1y=y+xxy=y+xxy=1x+1y\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}
Option E is correct
Note-Remember to recall the basic formulas of trigonometric functions while solving these types of questions. Note that tan⁡θ=1cot⁡θ\tan \theta = \dfrac{1}{{\cot \theta }} . Also we need to understand the basic concept and rearrangement that is done while solving this question.