If $$1$$−$$tanx1+tanx=tany$$ and x−$$y=$$π$$6$$, then x, y are respectively

Correct option is 4all the above

Questions

$I f \frac{1 - \tan x}{1 + \tan x} = \tan y a n d x - y = \frac{π}{6}, t h e n x, y a r e r e s p e c t i v e l y$

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5π24,π24

−7π24,−11π24

−115π24,−119π24

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detailed solution

Correct option is D

We have tanπ4−x=tany=tanx−π6 ∵x−y=π6 ∴π4−x=nπ+x−π6 ⇒5π12−nπ=2x⇒x=5−12nπ24 and y=x−π6Options (1), (2), (3) correspond to n = 0, 1, 10 respectively .

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If 1−tanx1+tanx=tany and x−y=π6, then x, y are respectively

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$I f \frac{1 - \tan x}{1 + \tan x} = \tan y a n d x - y = \frac{π}{6}, t h e n x, y a r e r e s p e c t i v e l y$