Question #4aae2

1 Answer
Feb 12, 2017

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)