Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2-4x+1=0$ $(\alpha >\beta ),$ then the value of $f(\alpha ,\beta )=\frac{{\beta }^3}{2}{\text{cosec}}^2\left(\frac{1}{2}{\tan }^{-1}\frac{\beta }{\alpha }\right)+\frac{{\alpha }^3}{2}{\sec }^2\left(\frac{1}{2}{\tan }^{-1}\frac{\alpha }{\beta }\right)$ is

Answer 1

$\alpha ,\beta$ are roots of the equation $x^2-4x+1=0,\alpha >\beta f(\alpha ,\beta )=\frac{{\beta }^3}{2{\sin }^2\left(\frac{1}{2}{\tan }^{-1}\frac{\beta }{\alpha }\right)}+\frac{{\alpha }^3}{2{\cos }^2\left(\frac{1}{2}{\tan }^{-1}\frac{\alpha }{\beta }\right)}=\frac{{\beta }^3}{1-\cos \left({\tan }^{-1}\frac{\beta }{\alpha }\right)}+\frac{{\alpha }^3}{1+\cos \left({\tan }^{-1}\frac{\alpha }{\beta }\right)}=\frac{{\beta }^3}{1-\cos \left({\cos }^{-1}\frac{\alpha }{\sqrt{{\alpha }^2+{\beta }^2}}\right)}+\frac{{\alpha }^3}{1+\cos \left({\cos }^{-1}\frac{\beta }{\sqrt{{\alpha }^2+{\beta }^2}}\right)}=\frac{{\beta }^3}{1-\frac{\alpha }{\sqrt{{\alpha }^2+{\beta }^2}}}+\frac{{\alpha }^3}{1+\frac{\beta }{\sqrt{{\alpha }^2+{\beta }^2}}}=\sqrt{{\alpha }^2+{\beta }^2}\{\frac{{\beta }^3}{\sqrt{{\alpha }^2+{\beta }^2}-\alpha }+\frac{{\alpha }^3}{\sqrt{{\alpha }^2+{\beta }^2}+\beta }\}=\sqrt{{\alpha }^2+{\beta }^2}\{\frac{{\beta }^3(\sqrt{{\alpha }^2+{\beta }^2}+\alpha )}{{\beta }^2}+\frac{{\alpha }^3(\sqrt{{\alpha }^2+{\beta }^2}-\beta )}{{\alpha }^2}\}=\sqrt{{\alpha }^2+{\beta }^2}(\beta \sqrt{{\alpha }^2+{\beta }^2}+\alpha \sqrt{{\alpha }^2+{\beta }^2})$

$=\sqrt{{\alpha }^2+{\beta }^2}\sqrt{{\alpha }^2+{\beta }^2}(\alpha +\beta )=({\alpha }^2+{\beta }^2)(\alpha +\beta )$
$=\left\{\right(\alpha +\beta {)}^2-2\alpha \beta \left\}\right(\alpha +\beta )$
Since, $\alpha +\beta =4,\alpha \beta =1,$
$f(\alpha ,\beta )=(16-2)\left(4\right)=14\times 4=56$