If x `sin^3 θ` `theta`  + y `cos^3 θ ``theta`  = sin θ cos θ, and x sin θ = y cos θ, prove that `x^2 + y^2 = 1` kindly give me a solution for the question  

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You need to substitute `y cos theta`  for `x sin^3 theta`  in equation x `sin^3 theta + y cos^3 theta = sin theta*cos theta`  such that:

`y*cos theta*sin^2 theta + y cos^3 theta = sin theta*cos theta`

You should factor out `y cos theta ` such that:

`y cos theta(sin^2...

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You need to substitute `y cos theta`  for `x sin^3 theta`  in equation x `sin^3 theta + y cos^3 theta = sin theta*cos theta`  such that:

`y*cos theta*sin^2 theta + y cos^3 theta = sin theta*cos theta`

You should factor out `y cos theta ` such that:

`y cos theta(sin^2 theta + cos^2 theta) = sin theta*cos theta`

You should remember that `sin^2 theta + cos^2 theta = 1`  such that:

`y cos theta = sin theta*cos theta`

Reducing by `cos theta`  yields:

`y = sin theta`

You should substitute `sin theta`  for y in equation `x sin theta = y cos theta`  such that:

`x sin theta = sin theta*cos theta`

`x = cos theta`

You should raise to square x and y such that:

`x^2 = cos^2 theta ; y^2 = sin^2 theta`

You need to add `x^2`  and `y^2`  such that:

`x^2 + y^2 = cos^2 theta + sin^2 theta`

By fundamental formula of trigonometry yields `cos^2 theta + sin^2 theta = 1` , hence `x^2 + y^2 = 1` .

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