Given that #x = cottheta+ tantheta# and #y = sec theta - costheta#, how do you find an expression for #x# and #y# in terms of #x# and #y# ?

Given:

#x=cot(theta) + tan(theta)#; #y=sec(theta)- cos(theta)#

Write both equation in terms of sine and cosine functions only:

#x=cos(theta)/sin(theta) + sin(theta)/cos(theta)#; #y=1/cos(theta)- cos(theta)#

Make common denominators for both equations:

#x=cos^2(theta)/(sin(theta)cos(theta)) + sin^2(theta)/(sin(theta)cos(theta))#; #y=1/cos(theta)- cos^2(theta)/cos(theta)#

Combine both equations over their respective common denominators:

#x=(cos^2(theta) + sin^2(theta))/(sin(theta)cos(theta))#; #y=(1- cos^2(theta))/cos(theta)#

We know that #cos^2(theta) + sin^2(theta) = 1# and #1- cos^2(theta)= sin^2(theta)#:

#x=1/(sin(theta)cos(theta))#; #y=sin^2(theta)/cos(theta)#

Divide y by x:

#y/x = (sin^2(theta)/cos(theta))/(1/(sin(theta)cos(theta)))#

#y/x = sin^3(theta)#

#sin(theta) = root(3)(y/x)#

#theta = sin^-1(root(3)(y/x))#

Substitute into the original equations:

#x=cot(sin^-1(root(3)(y/x))) + tan(sin^-1(root(3)(y/x)))#; #y=sec(sin^-1(root(3)(y/x)))- cos(sin^-1(root(3)(y/x)))#

A little different answer. Rewriting in cosine and sine:

#x = costheta/sintheta + sintheta/costheta#

#x= (cos^2theta + sin^2theta)/(costhetasintheta)#

Recalling that #cos^2theta + sin^2theta = 1#:

#x= 1/(costhetasintheta)#

#costhetasintheta = 1/x#

We know that #2sinxcosx = sin(2x)#, thus:

#1/2sin(2theta) = 1/x#

#sin(2theta) = 2/x#

#2theta = arcsin(2/x)#

#theta= 1/2arcsin(2/x)#

Therefore, substituting:

#y = sec(1/2arcsin(2/x)) - cos(1/2arcsin(2/x))#

Hopefully this helps!