Question

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

Answer 1

$f \left(\alpha, \beta\right)=\frac{\beta^{3}}{2} cosec^{2} \left(\frac{1}{2} tan^{-1}\frac{\beta}{\alpha}\right)+\frac{\alpha^{3}}{2} sec^{2} \left(\frac{1}{2} tan^{-1}\frac{\alpha}{\beta}\right)$

Let $tan^{-1} \left(\frac{\beta}{\alpha}\right)=\theta$ and $tan^{-1}\left(\frac{\alpha}{\beta}\right)=\phi$
$f \left(\alpha,\beta\right)=\frac{\beta^{3}}{2\,sin^{2} \frac{\theta}{2}}+\frac{\alpha^{3}}{2\,cos^{2} \frac{\phi}{2}}=\frac{\beta^{3}}{1-cos\,\theta}+\frac{\alpha^{3}}{1+cos\,\phi} $
$=\frac{\beta^{3}}{1-\frac{\alpha}{\sqrt{\alpha^{2}+\beta^{2}}}}+\frac{\alpha^{3}}{1+\frac{\beta}{\sqrt{\alpha^{2}+\beta^{2}}}}$
$=\sqrt{\alpha^{2}+\beta^{2}} \left[\frac{\beta^{3}}{\sqrt{\alpha^{2}+\beta^{2}}-\alpha}+\frac{\alpha^{3}}{\sqrt{\alpha^{2}+\beta^{2} }+\beta}\right]$
$=\sqrt{\alpha^{2}+\beta^{2}}\left[\frac{\beta^{3}\left(\sqrt{\alpha^{2}+\beta^{2}}+\alpha\right)}{\beta^{2}}+\frac{\alpha^{3}\left(\sqrt{\alpha^{2}+\beta^{2}-\beta}\right)}{\alpha^{2}}\right]$
$=\sqrt{\alpha^{2}+\beta^{2}} \left[\beta \sqrt{\alpha^{2}+\beta^{2}}+\alpha\sqrt{\alpha^{2}+\beta^{2}}\right]$
$f \left(\alpha, \beta\right)=\left(\alpha^{2}+\beta^{2}\right)\left(\alpha+\beta\right)$
Now, $\alpha+\beta =4$ and $\alpha\,\beta=1$
$f \left(\alpha,\beta\right)=\left(\left(\alpha+\beta\right)^{2}-2\alpha\beta)\right) \left(\alpha+\beta\right)$
$=\left(16-2\right)\left(4\right)=56$