Question

Question 49) If tan A-tan B =X
and cot B-cot A=y,then cot
(A-B)= ? *
O 1/y - 1/2
O 1/x-1/y
O 1/x + 1/y
Oxy /x-y​

Answer 1

Answer:

$\blue{\mathfrak{\underline{\large{Given}}}:} \\ tanA-tanB=x \\ cotB-cotA=y \\ \blue{\mathfrak{\underline{\large{To \: find}}}:} \\ cot(A-B)=? \\ \blue{\mathfrak{\underline{\large{Finding}}}:} \\ cotB-cotA=y - (i) \\ tanA-tanB=x-(ii) \\ \blue{\mathfrak{\underline{\large{solving \: equation \: (ii)}}}:}\\ \frac{1}{cotA} - \frac{1}{cotB} = x \\ \frac{cotB-cotA}{cotA.cotB} =x \\ \blue{\mathfrak{\underline{\large{By \: equation \:(i)}}}:} \\ \frac{y}{cotA.cotB} =x \\ cotA.cotB= \frac{y}{x} - (iii) \\ \blue{\mathfrak{\large{\blue{\mathfrak{\underline{\large{Using \: formula}}}:}}}} \\ cot(A-B) = \frac{cotA.cotB+1}{cotB-cotA} \\ \blue{\mathfrak{\underline{\large{Put \: the \: value \: of \: equation \: (i) \: and \: (iii) \: in \: formula}}}:} \\ cot(A-B)= \frac{ \frac{y}{x} + 1}{y} \\ = \frac{ \frac{y + x}{x} }{y} \\ = \frac{y + x}{xy} \\ = \frac{y}{xy} + \frac{x}{xy} \\ = \frac{1}{x} + \frac{1}{y} \\ \blue{\rightarrow \:(c) \: is \: correct \: answer} \\ \red{\mathfrak{\underline{\large{Hope \: It \: Helps \: You }}}} \\ \green{\mathfrak{\underline{\large{Mark \: Me \: Brainliest}}}}$