Question

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` ?

Answer 1

`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)))`

Explanation:

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)))`

Answer 2

`y = sec(1/2arcsin(2/x)) - cos(1/2arcsin(2/x))`
`x = cot(1/2arcsin(2/x)) + tan(1/2arcsin(2/x))`

Explanation:

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!