A hyperbola having the transverse axis of length 2sinθ , is confocal with ellipse 3x²+4y²-12=0 then its equation is ____________The answer is x²cosec² θ- y²sec² θ = 1 please explain me how

Ellipse equation is 3x² + 4y² - 12 = 0 
Hence $\frac{x^2}{4} +\frac{y^2}{3} = 1$
So a² = 4 and b² = 3
So eccentricity = $\sqrt{1-\frac{b^2}{a^2}} =\pm \frac{1}{2}$
So focus of the ellipse = (ae ,0)  and (-ae ,0 )
Hence the focus = (1,0) and ( -1 ,0 )

As the hyperbola is confocal , so focus is same , hence C for hyperbola = 1
And length of transverse axis = 2sin$\theta$
So length of semi transverse axis  = A = sin$\theta$ 
And C² = A² + B²  [ here A , B and C are parameters in hyperbola , similar to ellipse ]
So B² = 1- sin²$\theta$  = cos²$\theta$ 

Hence the equation  of hyperbola is $\frac{x^2}{A^2} -\frac{y^2}{B^2} = 1 or x^2\cos e{c}^2\theta - y^{2}se{c}^2\theta = 1$