The correct options are
A sinθcosθ=1x
B sinθtanθ=y
C (x2y)23−(xy2)23=1
Since cotθ+tanθ=x and
secθ−cosθ=y
option (A):
∴1x=1cotθ+tanθ=sinθcosθcos2θ+sin2θ
=sinθcosθ1
∴sinθcosθ=1x
option (B):
since, y=secθ−cosθ
=1−cos2θcosθ=sin2θcosθ
sinθtanθ
Hence, sinθtanθ=y
option (C):
since, x2y
=1sin2θcos2θ×sin2θcosθ=1cos3θ
⇒(1x2y)23=cos2θ
⇒(x2y)23=sec2θ....(i)
and xy2=1sinθcosθ×sin4θcos2θ=tan3θ
⇒(xy2)23=tan2θ....(ii)
From equation (i) and (ii),
(x2y)23−(xy2)23=1