Question

If sin(x+y)=cos(x+y) then prove tanx=1-tany/1+tany

Answer 1

Given

$\sf sin(x+y)=cos(x+y)$

To prove

$\sf tanx=\dfrac{1-tany}{1+tany}$

Explanation

$= > sin(x + y) = cos(x + y)$

$= > \dfrac{ sin(x + y)}{cos(x + y)} = 1$

$= > tan(x + y) = 1$

Expanding using the formula

$\underline{ \sf \: tan(a + b) = \dfrac{tana + tanb}{1 - tanatanb} }$

then

$= > \dfrac{tanx + tany}{1 - tanxtany} = 1$

$tanx + tany = 1 - tanxtany$

$= > tanx + tanxtany = 1 - tany$

$= > tanx(1 + tany) = 1 - tany$

$= > tanx = \dfrac{1 - tany}{1 + tany}$

$\boxed{ \sf \: tanx = \dfrac{1 - tany}{1 + tany} }$

Hence proved

Some more important formulas

$= > sin(x + y) = sinxcosy + cosxsiny$

$= > cos(x + y) = cosxcosy - sinxsiny$

$= > tan( x - y) = \dfrac{tanx - tany}{1 + tanxtany}$

$=>cot(x+y)=\dfrac{cotxcoty-1}{coty+cotx}$

$=>cot(x-y)=\dfrac{cotxcoty+1}{coty-cotx}$