Question

tanA-tanB=m cotA-cotB=n prove that cot(A-B)=1/m-1/n​

Answer 1

Answer:

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Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

$cot(A - B )= \displaystyle \: \frac{cotA \: cotB \: + 1}{ cotB -cotA }$

GIVEN

$tanA-tanB=m \: \: and \: \: cotA-cotB=n$

TO PROVE

$\displaystyle \: cot(A-B) \: = \frac{1}{m} - \frac{1}{n}$

PROOF

RHS

$\displaystyle \: \frac{1}{m} - \frac{1}{n}$

$= \displaystyle \: \frac{1}{tanA-tanB} - \frac{1}{cotA-cotB}$

$= \displaystyle \: \frac{1}{ \frac{1}{cotA} - \frac{1}{cotB} } - \frac{1}{cotA-cotB}$

$= \displaystyle \: \frac{cotA \: cotB}{ cotB -cotA } - \frac{1}{cotA-cotB}$

$= \displaystyle \: \frac{cotA \: cotB}{ cotB -cotA } + \frac{1}{cotB -cotA }$

$= \displaystyle \: \frac{cotA \: cotB \: + 1}{ cotB -cotA }$

$= \displaystyle \: cot(A - B) \:$

= LHS

Hence proved