Question #4aae2

Question

1303 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-12T15:03:35

#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)#