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Question: If \(\tan A - \tan B = x\) and \(\cot B - \cot A = y,\) then \(\cot(A - B) =\)...

If tan⁡A−tan⁡B=x\tan A - \tan B = x and cot⁡B−cot⁡A=y,\cot B - \cot A = y, then cot⁡(A−B)=\cot(A - B) =

A

1x+y\frac{1}{x} + y

B

1xy\frac{1}{xy}

C

1x−1y\frac{1}{x} - \frac{1}{y}

D

1x+1y\frac{1}{x} + \frac{1}{y}

Answer

1x+1y\frac{1}{x} + \frac{1}{y}

Explanation

Solution

cot⁡(A−B)=1tan⁡(A−B)=1+tan⁡Atan⁡Btan⁡A−tan⁡B\cot(A - B) = \frac{1}{\tan(A - B)} = \frac{1 + \tan A\tan B}{\tan A - \tan B}

=1tan⁡A−tan⁡B+tan⁡Atan⁡Btan⁡A−tan⁡B=1x+1y= \frac{1}{\tan A - \tan B} + \frac{\tan A\tan B}{\tan A - \tan B} = \frac{1}{x} + \frac{1}{y}.