Question

tan x-tan y
15. Prove that tan(x - y) =
1+tan x tan y/1+tan x tan y

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Answer 1

$\tt \fcolorbox{skyblue}{skyblue}{I \: think \: GIVEN \: is \: tan \: x \: - \: tan y = \: 1 + tan x tan y}$

PROOF :

$\sf \: We \: \: know,$

$\mapsto \: \: \: \sf \sin (x - y) = \sin \: x \cos \: y - \cos \: x \sin \: y$

$\mapsto \: \: \sf \: \cos(x - y) = \cos \: x \cos \: y + \sin \: x \sin \: y$

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$\sf \boxed {\tt \large{L \: H \: S}} \\ \\ \sf \tan \: (x - y) \\ \\ \implies \: \sf \frac{ \sin \: (x - y)}{ \cos \: (x - y)} \\ \\ \implies \sf \frac{ \sin \: x \cos \: y - \cos \: x \sin \: y}{ \cos \: x \cos \: y + \sin \: x \cos \: y} \\ \\ \implies \: \sf \frac{ \sin \: x \cos \: y - \cos \: x \sin \: y}{ \cos \: x \cos \: y + \sin \: x \cos \: y} \times \frac{ \frac{1}{ \cos \: x \cos \: y} }{ \frac{1}{ \cos \: x \cos \: y} } \\ \\ \implies \sf \: \frac{ \tan \: x \: - \tan \: y}{1 + \tan \: x \tan \: y} \\ \\ \\ \sf \therefore \boxed {\tt \large{R\: \: H\: \: S}}$

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HOPE THIS IS HELPFUL...