By writing the first equation as a single fraction we can see that

x = (sin²a + cos²a)/(sin a cos a) = 1/(sin a cos a),

For the second equation we can see that

y = (2– 2cos²a)/cos a = 2 sin²a/cos a

Therefore y/x = 2sin³a , or y/2x = sin³a

and x²y = 1/(sin a cos a)² x 2sin²a/cos a = 2/cos³a, so 2/x²y = cos³a

This means that (y/2x)^(2/3) = (sin³ a)^(2/3) =sin²a, and (2/x²y)^(2/3)= cos²a.

Using the fact that sin²a + cos²a = 1 for any a (as we did earlier), we obtain

(y/2x)^(2/3) + (2/x²y)^(2/3) = 1

Clearing the denominators, by multiplying through by 2^2/3x^4/3y^2/3 , we get

y^4/3x^2/3 + 2^4/3 = 2^2/3x^4/3y^2/3

or

(y²x)^2/3 + 2^4/3 = (2x²y)^2/3

Now we will cube both sides of the equation:

(y²x)² + 3(2^4/3)(y²x)^4/3 + 3(2^8/3)(y²x)^2/3 + 4²= (2x²y)²

Now for the very sneaky bit! Notice that this is the same as

(y²x)² + 3(2^4/3)(y²x)^2/3((y²x)^2/3 + 2^4/3) + 16 = 4x⁴y²

But recall from earlier that

(y²x)^2/3 + 2^4/3 = (2x²y)^2/3 so we can substitute for (y²x)^2/3 + 2^4/3, and then simplify

(y²x)² + 3(2^4/3)(y²x)^2/3(2x²y)^2/3 + 16 = 4x⁴y²

y⁴x²+ 3(2²y²x²) + 16 = 4x⁴y²

Simplifying again,

y⁴x²+ 12y²x² + 16 = 4x⁴y²

which is the same as our target expression:

4x⁴y² = 16 + x²y⁴ + 12x²y²

Challenge 284 : Tangled Trig