Question

Prove the trigonometric identity 1-sen(2y) / 1+ sen(2y) = tan ( π/4-y) / tan (π/4 + y)​

Answer 1

`\cfrac{1-sin(2y)}{1+sin(2y)}~~ = ~~\cfrac{tan\left( (\pi )/(4)-y \right)}{tan\left( (\pi )/(4)+y \right)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \boxed{tan\left( (\pi )/(4)-y \right)}\implies \cfrac{tan\left( (\pi )/(4) \right)-tan(y)}{1+tan\left( (\pi )/(4) \right)tan(y)}\implies \cfrac{1-tan(y)}{1+1\cdot tan(y)}`

`\cfrac{1-(sin(y))/(cos(y))}{1+(sin(y))/(cos(y))}\implies \cfrac{~~ ( cos(y)-sin(y))/(cos(y) ) ~~}{( cos(y)+sin(y))/(cos(y) )}\implies \cfrac{ cos(y)-sin(y)}{ cos(y)+sin(y)} \\\\[-0.35em] ~\dotfill\\\\ \boxed{tan\left( (\pi )/(4)+y \right)}\implies \cfrac{tan\left( (\pi )/(4) \right)+tan(y)}{1-tan\left( (\pi )/(4) \right)tan(y)}\implies \cfrac{1+tan(y)}{1-1\cdot tan(y)}`

`\cfrac{1+(sin(y))/(cos(y))}{1-(sin(y))/(cos(y))}\implies \cfrac{~~ ( cos(y)+sin(y))/(cos(y) ) ~~}{( cos(y)-sin(y))/(cos(y) )}\implies \cfrac{ cos(y)+sin(y)}{ cos(y)-sin(y)} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{tan\left( (\pi )/(4)-y \right)}{tan\left( (\pi )/(4)+y \right)}\implies \cfrac{ ~~ ( cos(y)-sin(y))/( cos(y)+sin(y)) ~~ }{( cos(y)+sin(y))/( cos(y)-sin(y))}\implies \cfrac{ (~~cos(y)-sin(y)~~)^2}{ (~~cos(y)+sin(y)~~)^2}`

`\cfrac{cos^2(y)-2sin(y)cos(y)+sin^2(y)}{cos^2(y)+2sin(y)cos(y)+sin^2(y)}\implies \cfrac{cos^2(y)+sin^2(y)-2sin(y)cos(y)}{cos^2(y)+sin^2(y)+2sin(y)cos(y)} \\\\\\ \cfrac{1-2sin(y)cos(y)}{1+2sin(y)cos(y)}\implies \boxed{\cfrac{1-sin(2y)}{1+sin(2y)}}`