If x=TanA-TanB, y=CotB-CotA then 1x+1y=

x=tan A-tan B y=1tan B-1tan A =tan A-tan Btan A tan B 1x+1y=1tan A-tan B+tan A tan Btan A-tan B =1+tan A tan Btan A-tan B =1tan (A-B) =cot(A-B)

If x=TanA-TanB, y=CotB-CotA then 1x+1y=

x=tan A-tan B y=1tan B-1tan A =tan A-tan Btan A tan B 1x+1y=1tan A-tan B+tan A tan Btan A-tan B =1+tan A tan Btan A-tan B =1tan (A-B) =cot(A-B)

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If $x = TanA - TanB, y = CotB - CotA$ then $\frac{1}{x} + \frac{1}{y} =$

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$Tan (B - A)$

$Tan (A - B)$

$Cot (B - A)$

$Cot (A - B)$

answer is A.

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Detailed Solution

$x = \tan A - \tan B y = \frac{1}{\tan B} - \frac{1}{\tan A} = \frac{\tan A - \tan B}{\tan A \tan B} \frac{1}{x} + \frac{1}{y} = \frac{1}{\tan A - \tan B} + \frac{\tan A \tan B}{\tan A - \tan B} = \frac{1 + \tan A \tan B}{\tan A - \tan B} = \frac{1}{\tan (A - B)} = \cot (A - B)$