Xsin(theta)-Ysin(theta)=sqrtX^2+Y^2 and cos^2(theta)/a^2+sin^2(theta)/b^2=1/X^2+Y^2 then find correct relation
xsinθ - ysinθ = √(x^2+y^2)
cos^2θ/a^2 + sin^2θ/b^2 = 1/(x^2+y^2)
not sure what you mean by the "correct relation" We can solve for θ or solve for x and y.
(x-y)^2 sin^2θ = x^2+y^2
so, sin^2θ = (x^2+y^2)/(x-y)^2 and we have
(1-((x^2+y^2)/(x-y)^2))/a^2 + ((x^2+y^2)/(x-y)^2)/b^2 = 1/(x^2+y^2)
I don't know how you want to massage that, but it will remain messy, I'm sure.
I suspect a typo, since you have sinθ twice in the first line. I'd expected a cosθ there somewhere.
To find the correct relation between the given equations, let's simplify and manipulate the equations step by step.
1. Starting with the first equation:
Xsin(theta) - Ysin(theta) = sqrt(X^2 + Y^2)
Factor out sin(theta):
sin(theta) (X - Y) = sqrt(X^2 + Y^2)
Divide both sides by (X - Y):
sin(theta) = sqrt(X^2 + Y^2) / (X - Y)
Take the square of both sides since (sin^2(theta) = 1 - cos^2(theta)):
sin^2(theta) = X^2 + Y^2 / (X - Y)^2
This simplifies to:
sin^2(theta) = (X^2 + Y^2) / (X^2 - 2XY + Y^2)
2. Moving on to the second equation:
cos^2(theta)/a^2 + sin^2(theta)/b^2 = 1/(X^2 + Y^2)
Multiply both sides by a^2(X^2 + Y^2):
(cos^2(theta)(X^2 + Y^2))/a^2 + (sin^2(theta)(X^2 + Y^2))/b^2 = 1
Distribute the terms:
(X^2cos^2(theta) + Y^2cos^2(theta))/a^2 + (X^2sin^2(theta) + Y^2sin^2(theta))/b^2 = 1
Since sin^2(theta) = 1 - cos^2(theta), substitute in:
(X^2cos^2(theta) + Y^2cos^2(theta))/a^2 + (X^2(1 - cos^2(theta)) + Y^2(1 - cos^2(theta)))/b^2 = 1
Now simplify:
(X^2cos^2(theta) + Y^2cos^2(theta))/a^2 + (X^2 - X^2cos^2(theta) + Y^2 - Y^2cos^2(theta))/b^2 = 1
Combine like terms:
(X^2(cos^2(theta) - cos^2(theta)) + Y^2(cos^2(theta) - cos^2(theta)) + X^2 + Y^2)/a^2b^2 = 1
Simplifying further:
(X^2 + Y^2)/a^2b^2 = 1
This equation can be written as:
X^2 + Y^2 = a^2b^2
Now we have the equations:
sin^2(theta) = (X^2 + Y^2) / (X^2 - 2XY + Y^2)
X^2 + Y^2 = a^2b^2
From these two equations, we can conclude that the correct relation is:
sin^2(theta) = (X^2 + Y^2) / (X^2 - 2XY + Y^2) = a^2b^2 / (X^2 + Y^2)