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Answer 1 · 2017-10-25T20:57:36

#"See below"#.

#x = a /(cos^3 theta)#
#y = b sin^3theta/cos^3theta# then

#x/y = a/b sin^3 theta# and then

#theta =arcsin( root(3)((bx)/(ay)))#

Answer 2 · 2017-10-26T02:30:34

We have

#sec^3theta=x#

#=>sec^2theta=x^(2/3)......【1】#

And

#tan^3theta=y#

#=>tan^2theta=y^(2/3)......【2】#

So subtracting [2] from [1] we get

#x^(2/3)-y^(2/3)=sec^2theta-tan^2theta#

#=>x^(2/3)-y^(2/3)=1#