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Question

Mathematics Question on Trigonometric Functions

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

1y−1x\frac{1}{y} -\frac{1}{x}

B

1x−1y\frac{1}{x} - \frac{1}{y}

C

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

D

none of thesenone\, of\, these

Answer

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

Explanation

Solution

We have, tan⁡A−tan⁡B=x\tan A - \tan B = x and cot⁡B−cot⁡A=y\cot B - \cot A = y
⇒1tan⁡B−1tan⁡A=y\Rightarrow \frac{1}{\tan B } - \frac{1}{\tan A} =y
tan⁡A−tan⁡Btan⁡Atan⁡B=y⇒xy=tan⁡Atan⁡B\frac{\tan A -\tan B}{\tan A \tan B} = y \Rightarrow \frac{x}{y} = \tan A \tan B
cot⁡(A−B)=1tan⁡(A−B)=1tan⁡A−tan⁡B1+tan⁡Atan⁡B\cot\left(A-B\right)= \frac{1}{\tan \left(A-B\right) } = \frac{1}{\frac{\tan A -\tan B}{1 +\tan A \tan B}}
=1x1+xy=y+xxy=1x+1y= \frac{1}{\frac{x}{1+ \frac{x}{y}}} = \frac{y+x}{xy } = \frac{1}{x} + \frac{1}{y}