Question

Question #4aae2

Answer 1

`y=(Sin4x+cos4x)/(sin4x-cos4x)`

`=>y/1=(sin4x+cos4x)/(sin4x-cos4x)`

By componendo and dividendo

`=>(y+1)/(y-1)=(2sin4x)/(2cos4x)`

`=>(y+1)/(y-1)=tan4x`

`=>(y+1)/(y-1)=(2tan2x)/(1-tan^2 2x`

Again by componendo and dividendo

`=>(2y)/2=(2tan2x+1-tan^2 2x)/(2tan2x-1+tan^2 2x)`

`=>y=((4tanx)/(1-tan^2x)+1-(4tan^2x)/(1-tan^2x)^2)/((4tanx)/(1-tan^2x)-1+(4tan^2x)/(1-tan^2x)^2)`