If $\alpha$ and $\beta$ are the roots of the equation $x^2-4x+1=0$ then find the value of $$f(\alpha, \beta) =\frac {(\beta)^3}{2} \text{cosec}^2\left (\frac {1}{2}\arctan \frac {\beta}{\alpha}\right) +\frac {(\alpha)^3}{2} \sec^2\left (\frac {1}{2}\arctan \frac {\alpha}{\beta}\right)$$
My try :
I wrote $$\arctan \frac {\beta}{\alpha}=\theta$$ hence we get $$\arctan \frac {\alpha}{\beta}=\frac {\pi}{2}-\theta$$ Using this I tried to simplify the expression but the cubes are messing up the method i tried. Do this type of question have any other method. If not someone please provide a better way to solve the problem using my approach. Thanks.
As $\alpha,\beta>0$
Let $\arctan\dfrac\alpha\beta=u\implies0<u<\dfrac\pi2\tan u=\dfrac\alpha\beta,$
$\cos u=+\dfrac1{\sqrt{1+\left(\dfrac\alpha\beta\right)^2}}=\dfrac{\sqrt{\beta^2}}{\sqrt{\alpha^2+\beta^2}}=\dfrac{|\beta|}{\sqrt{\alpha^2+\beta^2}}$
As $\beta>0,|\beta|=+\beta$
Using $\cos2A=2\cos^2A-1=1-2\sin^2A,$
$$f(\alpha,\beta)=\dfrac{\beta^3}{1-\cos\left(\arctan\dfrac\beta\alpha\right)}+\dfrac{\alpha^3}{1+\cos\left(\arctan\dfrac\alpha\beta\right)}$$
$$=\dfrac{\beta^3\sqrt{\alpha^2+\beta^2}}{\sqrt{\alpha^2+\beta^2}-\alpha}+\dfrac{\alpha^3\sqrt{\alpha^2+\beta^2}}{\sqrt{\alpha^2+\beta^2}+\beta}$$
$$=\sqrt{\alpha^2+\beta^2}\cdot\sqrt{\alpha^2+\beta^2}(\alpha+\beta)=?$$