If cotθ+tanθ=x and secθ−cosθ=y, then
xsinθ⋅cosθ=1
sin2θ=ycosθ
x2y1/3+xy21/3=1
x2y2/3−xy22/3=1
cotθ+tanθ=x
or cosθsinθ+sinθcosθ=xor 1=xsinθcosθ
Now secθ−cosθ=y
1cosθ−cosθ=yor sin2θ=ycosθ
now x2y=1sin2θcos2θ⋅sin2θcosθ=1cos3θand xy2=sin3θcos3θ∴x2y2/3−xy22/3=1cos2θ−sin2θcos2θ=1