Question

If `cos^-1[(x^2 - y^2)/(x^2 + y^2)] = 2k    "show that y" dy/dx = xtan^2k`

Answer 1

`cos^-1[(x^2 - y^2)/(x^2 + y^2)] = 2k`

∴ `[ x^2 - y^2 ]/[ x^2 + y^2 ] = cos2k`

∴ `x^2 - y^2 = cos 2k ( x^2 + y^2 )`

Differentiate w.r.t. x.

`2x - 2y (dy/dx) = cos 2k (2x) + cos 2k (2y dy/dx)`

∴ 2x - 2x cos 2k = 2y cos 2k `dy/dx + 2y dy/dx`

∴ x - x cos 2k = `dy/dx` ( cos2k. y + y )

∴ `dy/dx = [ x( 1 - cos 2k )] / [ y( 1 + cos 2k )]`

∴ `dy/dx = [ x 2sin^2 k ]/[ y 2 cos^2k ]`

∴ `y (dy/dx) = xtan^2 k`